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csc^2(x)+2cos(x)+1=0

解

csc^{2} (x) + 2 cos (x) + 1 = 0

解

以下の解はない : x \in R

解答ステップ

csc^{2} (x) + 2 cos (x) + 1 = 0

両辺から

2 cos (x)

を引く

csc^{2} (x) + 1 = - 2 cos (x)

両辺を2乗する

(csc^{2} (x) + 1)^{2} = (- 2 cos (x))^{2}

両辺から

(- 2 cos (x))^{2}

を引く

(csc^{2} (x) + 1)^{2} - 4 cos^{2} (x) = 0

三角関数の公式を使用して書き換える

(1 + csc^{2} (x))^{2} - 4 cos^{2} (x)

基本的な三角関数の公式を使用する:

csc (x) = \frac{1}{sin ( x )}

= (1 + (\frac{1}{sin ( x )})^{2})^{2} - 4 cos^{2} (x)

(\frac{1}{sin ( x )})^{2} = \frac{1}{sin ^{2} ( x )}

(\frac{1}{sin ( x )})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sin ^{2} ( x )}

規則を適用

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( x )}

= (1 + \frac{1}{sin ^{2} ( x )})^{2} - 4 cos^{2} (x)

(1 + \frac{1}{sin ^{2} ( x )})^{2} - 4 cos^{2} (x) = 0

因数

(1 + \frac{1}{sin ^{2} ( x )})^{2} - 4 cos^{2} (x) : (1 + \frac{1}{sin ^{2} ( x )} + 2 cos (x)) (1 + \frac{1}{sin ^{2} ( x )} - 2 cos (x))

(1 + \frac{1}{sin ^{2} ( x )})^{2} - 4 cos^{2} (x)

(1 + \frac{1}{sin ^{2} ( x )})^{2} - 4 cos^{2} (x)

を書き換え

(1 + \frac{1}{sin ^{2} ( x )})^{2} - (2 cos (x))^{2}

(1 + \frac{1}{sin ^{2} ( x )})^{2} - 4 cos^{2} (x)

4

を書き換え

2^{2}

= (1 + \frac{1}{sin ^{2} ( x )})^{2} - 2^{2} cos^{2} (x)

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

2^{2} cos^{2} (x) = (2 cos (x))^{2}

= (1 + \frac{1}{sin ^{2} ( x )})^{2} - (2 cos (x))^{2}

= (1 + \frac{1}{sin ^{2} ( x )})^{2} - (2 cos (x))^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

(1 + \frac{1}{sin ^{2} ( x )})^{2} - (2 cos (x))^{2} = ((1 + \frac{1}{sin ^{2} ( x )}) + 2 cos (x)) ((1 + \frac{1}{sin ^{2} ( x )}) - 2 cos (x))

= ((1 + \frac{1}{sin ^{2} ( x )}) + 2 cos (x)) ((1 + \frac{1}{sin ^{2} ( x )}) - 2 cos (x))

改良

= (\frac{1}{sin ^{2} ( x )} + 2 cos (x) + 1) (\frac{1}{sin ^{2} ( x )} - 2 cos (x) + 1)

(1 + \frac{1}{sin ^{2} ( x )} + 2 cos (x)) (1 + \frac{1}{sin ^{2} ( x )} - 2 cos (x)) = 0

各部分を別個に解く

1 + \frac{1}{sin ^{2} ( x )} + 2 cos (x) = 0 or 1 + \frac{1}{sin ^{2} ( x )} - 2 cos (x) = 0

1 + \frac{1}{sin ^{2} ( x )} + 2 cos (x) = 0 :

解なし

1 + \frac{1}{sin ^{2} ( x )} + 2 cos (x) = 0

簡素化

1 + \frac{1}{sin ^{2} ( x )} + 2 cos (x) : \frac{sin ^{2} ( x ) + 1 + 2 sin ^{2} ( x ) cos ( x )}{sin ^{2} ( x )}

1 + \frac{1}{sin ^{2} ( x )} + 2 cos (x)

元を分数に変換する:

1 = \frac{1 sin ^{2} ( x )}{sin ^{2} ( x )}, 2 cos (x) = \frac{2 cos ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

= \frac{1 \cdot sin ^{2} ( x )}{sin ^{2} ( x )} + \frac{1}{sin ^{2} ( x )} + \frac{2 cos ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot sin ^{2} ( x ) + 1 + 2 cos ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

乗算:

1 \cdot sin^{2} (x) = sin^{2} (x)

= \frac{sin ^{2} ( x ) + 1 + 2 sin ^{2} ( x ) cos ( x )}{sin ^{2} ( x )}

\frac{sin ^{2} ( x ) + 1 + 2 sin ^{2} ( x ) cos ( x )}{sin ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin^{2} (x) + 1 + 2 sin^{2} (x) cos (x) = 0

三角関数の公式を使用して書き換える

1 + sin^{2} (x) + 2 cos (x) sin^{2} (x)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 + 1 - cos^{2} (x) + 2 cos (x) (1 - cos^{2} (x))

簡素化

1 + 1 - cos^{2} (x) + 2 cos (x) (1 - cos^{2} (x)) : 2 cos (x) - 2 cos^{3} (x) - cos^{2} (x) + 2

1 + 1 - cos^{2} (x) + 2 cos (x) (1 - cos^{2} (x))

拡張

2 cos (x) (1 - cos^{2} (x)) : 2 cos (x) - 2 cos^{3} (x)

2 cos (x) (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 2 cos (x), b = 1, c = cos^{2} (x)

= 2 cos (x) \cdot 1 - 2 cos (x) cos^{2} (x)

= 2 \cdot 1 \cdot cos (x) - 2 cos^{2} (x) cos (x)

簡素化

2 \cdot 1 \cdot cos (x) - 2 cos^{2} (x) cos (x) : 2 cos (x) - 2 cos^{3} (x)

2 \cdot 1 \cdot cos (x) - 2 cos^{2} (x) cos (x)

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 \cdot cos (x)

数を乗じる:

2 \cdot 1 = 2

= 2 cos (x)

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

2 cos^{2} (x) cos (x)

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

cos^{2} (x) cos (x) = cos^{2 + 1} (x)

= 2 cos^{2 + 1} (x)

数を足す:

2 + 1 = 3

= 2 cos^{3} (x)

= 2 cos (x) - 2 cos^{3} (x)

= 2 cos (x) - 2 cos^{3} (x)

= 1 + 1 - cos^{2} (x) + 2 cos (x) - 2 cos^{3} (x)

数を足す:

1 + 1 = 2

= 2 cos (x) - 2 cos^{3} (x) - cos^{2} (x) + 2

= 2 cos (x) - 2 cos^{3} (x) - cos^{2} (x) + 2

2 - cos^{2} (x) + 2 cos (x) - 2 cos^{3} (x) = 0

置換で解く

2 - cos^{2} (x) + 2 cos (x) - 2 cos^{3} (x) = 0

仮定:

cos (x) = u

2 - u^{2} + 2 u - 2 u^{3} = 0

2 - u^{2} + 2 u - 2 u^{3} = 0 : u \approx 1.14213 \dots

2 - u^{2} + 2 u - 2 u^{3} = 0

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 2 u^{3} - u^{2} + 2 u + 2 = 0

ニュートン・ラプソン法を使用して

- 2 u^{3} - u^{2} + 2 u + 2 = 0

の解を1つ求める:

u \approx 1.14213 \dots

- 2 u^{3} - u^{2} + 2 u + 2 = 0

ニュートン・ラプソン概算の定義

f (u) = - 2 u^{3} - u^{2} + 2 u + 2

発見する

f^{^{'}} (u) : - 6 u^{2} - 2 u + 2

\frac{d}{d u} (- 2 u^{3} - u^{2} + 2 u + 2)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (2 u^{3}) - \frac{d}{d u} (u^{2}) + \frac{d}{d u} (2 u) + \frac{d}{d u} (2)

\frac{d}{d u} (2 u^{3}) = 6 u^{2}

\frac{d}{d u} (2 u^{3})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{3})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 3 u^{3 - 1}

簡素化

= 6 u^{2}

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

簡素化

= 2 u

\frac{d}{d u} (2 u) = 2

\frac{d}{d u} (2 u)

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d u}{d u}

共通の導関数を適用:

\frac{d u}{d u} = 1

= 2 \cdot 1

簡素化

= 2

\frac{d}{d u} (2) = 0

\frac{d}{d u} (2)

定数の導関数:

\frac{d}{d x} (a) = 0

= 0

= - 6 u^{2} - 2 u + 2 + 0

簡素化

= - 6 u^{2} - 2 u + 2

仮定:

u_{0} = - 1

Δ u_{n + 1} <

になるまで

u_{n + 1}

を計算する

0.000001

u_{1} = - 0.5 : Δ u_{1} = 0.5

f (u_{0}) = - 2 (- 1)^{3} - (- 1)^{2} + 2 (- 1) + 2 = 1

f^{^{'}} (u_{0}) = - 6 (- 1)^{2} - 2 (- 1) + 2 = - 2

u_{1} = - 0.5

Δ u_{1} = ∣ - 0.5 - (- 1) ∣ = 0.5

Δ u_{1} = 0.5

u_{2} = - 1.16666 \dots : Δ u_{2} = 0.66666 \dots

f (u_{1}) = - 2 (- 0.5)^{3} - (- 0.5)^{2} + 2 (- 0.5) + 2 = 1

f^{^{'}} (u_{1}) = - 6 (- 0.5)^{2} - 2 (- 0.5) + 2 = 1.5

u_{2} = - 1.16666 \dots

Δ u_{2} = ∣ - 1.16666 \dots - (- 0.5) ∣ = 0.66666 \dots

Δ u_{2} = 0.66666 \dots

u_{3} = - 0.78019 \dots : Δ u_{3} = 0.38647 \dots

f (u_{2}) = - 2 (- 1.16666 \dots)^{3} - (- 1.16666 \dots)^{2} + 2 (- 1.16666 \dots) + 2 = 1.48148 \dots

f^{^{'}} (u_{2}) = - 6 (- 1.16666 \dots)^{2} - 2 (- 1.16666 \dots) + 2 = - 3.83333 \dots

u_{3} = - 0.78019 \dots

Δ u_{3} = ∣ - 0.78019 \dots - (- 1.16666 \dots) ∣ = 0.38647 \dots

Δ u_{3} = 0.38647 \dots

u_{4} = 7.72232 \dots : Δ u_{4} = 8.50251 \dots

f (u_{3}) = - 2 (- 0.78019 \dots)^{3} - (- 0.78019 \dots)^{2} + 2 (- 0.78019 \dots) + 2 = 0.78072 \dots

f^{^{'}} (u_{3}) = - 6 (- 0.78019 \dots)^{2} - 2 (- 0.78019 \dots) + 2 = - 0.09182 \dots

u_{4} = 7.72232 \dots

Δ u_{4} = ∣ 7.72232 \dots - (- 0.78019 \dots) ∣ = 8.50251 \dots

Δ u_{4} = 8.50251 \dots

u_{5} = 5.12779 \dots : Δ u_{5} = 2.59452 \dots

f (u_{4}) = - 2 \cdot 7.72232 \dots^{3} - 7.72232 \dots^{2} + 2 \cdot 7.72232 \dots + 2 = - 963.21890 \dots

f^{^{'}} (u_{4}) = - 6 \cdot 7.72232 \dots^{2} - 2 \cdot 7.72232 \dots + 2 = - 371.25003 \dots

u_{5} = 5.12779 \dots

Δ u_{5} = ∣ 5.12779 \dots - 7.72232 \dots ∣ = 2.59452 \dots

Δ u_{5} = 2.59452 \dots

u_{6} = 3.41896 \dots : Δ u_{6} = 1.70882 \dots

f (u_{5}) = - 2 \cdot 5.12779 \dots^{3} - 5.12779 \dots^{2} + 2 \cdot 5.12779 \dots + 2 = - 283.70151 \dots

f^{^{'}} (u_{5}) = - 6 \cdot 5.12779 \dots^{2} - 2 \cdot 5.12779 \dots + 2 = - 166.02106 \dots

u_{6} = 3.41896 \dots

Δ u_{6} = ∣ 3.41896 \dots - 5.12779 \dots ∣ = 1.70882 \dots

Δ u_{6} = 1.70882 \dots

u_{7} = 2.31481 \dots : Δ u_{7} = 1.10414 \dots

f (u_{6}) = - 2 \cdot 3.41896 \dots^{3} - 3.41896 \dots^{2} + 2 \cdot 3.41896 \dots + 2 = - 82.78202 \dots

f^{^{'}} (u_{6}) = - 6 \cdot 3.41896 \dots^{2} - 2 \cdot 3.41896 \dots + 2 = - 74.97378 \dots

u_{7} = 2.31481 \dots

Δ u_{7} = ∣ 2.31481 \dots - 3.41896 \dots ∣ = 1.10414 \dots

Δ u_{7} = 1.10414 \dots

u_{8} = 1.63810 \dots : Δ u_{8} = 0.67671 \dots

f (u_{7}) = - 2 \cdot 2.31481 \dots^{3} - 2.31481 \dots^{2} + 2 \cdot 2.31481 \dots + 2 = - 23.53607 \dots

f^{^{'}} (u_{7}) = - 6 \cdot 2.31481 \dots^{2} - 2 \cdot 2.31481 \dots + 2 = - 34.77990 \dots

u_{8} = 1.63810 \dots

Δ u_{8} = ∣ 1.63810 \dots - 2.31481 \dots ∣ = 0.67671 \dots

Δ u_{8} = 0.67671 \dots

u_{9} = 1.28138 \dots : Δ u_{9} = 0.35671 \dots

f (u_{8}) = - 2 \cdot 1.63810 \dots^{3} - 1.63810 \dots^{2} + 2 \cdot 1.63810 \dots + 2 = - 6.19847 \dots

f^{^{'}} (u_{8}) = - 6 \cdot 1.63810 \dots^{2} - 2 \cdot 1.63810 \dots + 2 = - 17.37647 \dots

u_{9} = 1.28138 \dots

Δ u_{9} = ∣ 1.28138 \dots - 1.63810 \dots ∣ = 0.35671 \dots

Δ u_{9} = 0.35671 \dots

u_{10} = 1.15779 \dots : Δ u_{10} = 0.12358 \dots

f (u_{9}) = - 2 \cdot 1.28138 \dots^{3} - 1.28138 \dots^{2} + 2 \cdot 1.28138 \dots + 2 = - 1.28712 \dots

f^{^{'}} (u_{9}) = - 6 \cdot 1.28138 \dots^{2} - 2 \cdot 1.28138 \dots + 2 = - 10.41447 \dots

u_{10} = 1.15779 \dots

Δ u_{10} = ∣ 1.15779 \dots - 1.28138 \dots ∣ = 0.12358 \dots

Δ u_{10} = 0.12358 \dots

u_{11} = 1.14237 \dots : Δ u_{11} = 0.01542 \dots

f (u_{10}) = - 2 \cdot 1.15779 \dots^{3} - 1.15779 \dots^{2} + 2 \cdot 1.15779 \dots + 2 = - 0.12893 \dots

f^{^{'}} (u_{10}) = - 6 \cdot 1.15779 \dots^{2} - 2 \cdot 1.15779 \dots + 2 = - 8.35854 \dots

u_{11} = 1.14237 \dots

Δ u_{11} = ∣ 1.14237 \dots - 1.15779 \dots ∣ = 0.01542 \dots

Δ u_{11} = 0.01542 \dots

u_{12} = 1.14213 \dots : Δ u_{12} = 0.00023 \dots

f (u_{11}) = - 2 \cdot 1.14237 \dots^{3} - 1.14237 \dots^{2} + 2 \cdot 1.14237 \dots + 2 = - 0.00188 \dots

f^{^{'}} (u_{11}) = - 6 \cdot 1.14237 \dots^{2} - 2 \cdot 1.14237 \dots + 2 = - 8.11480 \dots

u_{12} = 1.14213 \dots

Δ u_{12} = ∣ 1.14213 \dots - 1.14237 \dots ∣ = 0.00023 \dots

Δ u_{12} = 0.00023 \dots

u_{13} = 1.14213 \dots : Δ u_{13} = 5.21652 E - 8

f (u_{12}) = - 2 \cdot 1.14213 \dots^{3} - 1.14213 \dots^{2} + 2 \cdot 1.14213 \dots + 2 = - 4.2312 E - 7

f^{^{'}} (u_{12}) = - 6 \cdot 1.14213 \dots^{2} - 2 \cdot 1.14213 \dots + 2 = - 8.11116 \dots

u_{13} = 1.14213 \dots

Δ u_{13} = ∣ 1.14213 \dots - 1.14213 \dots ∣ = 5.21652 E - 8

Δ u_{13} = 5.21652 E - 8

u \approx 1.14213 \dots

長除法を適用する:

\frac{- 2 u ^{3} - u ^{2} + 2 u + 2}{u - 1.14213 \dots} = - 2 u^{2} - 3.28427 \dots u - 1.75110 \dots

- 2 u^{2} - 3.28427 \dots u - 1.75110 \dots \approx 0

ニュートン・ラプソン法を使用して

- 2 u^{2} - 3.28427 \dots u - 1.75110 \dots = 0

の解を1つ求める:

以下の解はない:

u \in R

- 2 u^{2} - 3.28427 \dots u - 1.75110 \dots = 0

ニュートン・ラプソン概算の定義

f (u) = - 2 u^{2} - 3.28427 \dots u - 1.75110 \dots

発見する

f^{^{'}} (u) : - 4 u - 3.28427 \dots

\frac{d}{d u} (- 2 u^{2} - 3.28427 \dots u - 1.75110 \dots)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (3.28427 \dots u) - \frac{d}{d u} (1.75110 \dots)

\frac{d}{d u} (2 u^{2}) = 4 u

\frac{d}{d u} (2 u^{2})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{2})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 2 u^{2 - 1}

簡素化

= 4 u

\frac{d}{d u} (3.28427 \dots u) = 3.28427 \dots

\frac{d}{d u} (3.28427 \dots u)

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 3.28427 \dots \frac{d u}{d u}

共通の導関数を適用:

\frac{d u}{d u} = 1

= 3.28427 \dots \cdot 1

簡素化

= 3.28427 \dots

\frac{d}{d u} (1.75110 \dots) = 0

\frac{d}{d u} (1.75110 \dots)

定数の導関数:

\frac{d}{d x} (a) = 0

= 0

= - 4 u - 3.28427 \dots - 0

簡素化

= - 4 u - 3.28427 \dots

仮定:

u_{0} = - 1

Δ u_{n + 1} <

になるまで

u_{n + 1}

を計算する

0.000001

u_{1} = - 0.34775 \dots : Δ u_{1} = 0.65224 \dots

f (u_{0}) = - 2 (- 1)^{2} - 3.28427 \dots (- 1) - 1.75110 \dots = - 0.46682 \dots

f^{^{'}} (u_{0}) = - 4 (- 1) - 3.28427 \dots = 0.71572 \dots

u_{1} = - 0.34775 \dots

Δ u_{1} = ∣ - 0.34775 \dots - (- 1) ∣ = 0.65224 \dots

Δ u_{1} = 0.65224 \dots

u_{2} = - 0.79716 \dots : Δ u_{2} = 0.44940 \dots

f (u_{1}) = - 2 (- 0.34775 \dots)^{2} - 3.28427 \dots (- 0.34775 \dots) - 1.75110 \dots = - 0.85083 \dots

f^{^{'}} (u_{1}) = - 4 (- 0.34775 \dots) - 3.28427 \dots = - 1.89323 \dots

u_{2} = - 0.79716 \dots

Δ u_{2} = ∣ - 0.79716 \dots - (- 0.34775 \dots) ∣ = 0.44940 \dots

Δ u_{2} = 0.44940 \dots

u_{3} = - 5.02192 \dots : Δ u_{3} = 4.22476 \dots

f (u_{2}) = - 2 (- 0.79716 \dots)^{2} - 3.28427 \dots (- 0.79716 \dots) - 1.75110 \dots = - 0.40393 \dots

f^{^{'}} (u_{2}) = - 4 (- 0.79716 \dots) - 3.28427 \dots = - 0.09561 \dots

u_{3} = - 5.02192 \dots

Δ u_{3} = ∣ - 5.02192 \dots - (- 0.79716 \dots) ∣ = 4.22476 \dots

Δ u_{3} = 4.22476 \dots

u_{4} = - 2.89752 \dots : Δ u_{4} = 2.12439 \dots

f (u_{3}) = - 2 (- 5.02192 \dots)^{2} - 3.28427 \dots (- 5.02192 \dots) - 1.75110 \dots = - 35.69719 \dots

f^{^{'}} (u_{3}) = - 4 (- 5.02192 \dots) - 3.28427 \dots = 16.80342 \dots

u_{4} = - 2.89752 \dots

Δ u_{4} = ∣ - 2.89752 \dots - (- 5.02192 \dots) ∣ = 2.12439 \dots

Δ u_{4} = 2.12439 \dots

u_{5} = - 1.81080 \dots : Δ u_{5} = 1.08672 \dots

f (u_{4}) = - 2 (- 2.89752 \dots)^{2} - 3.28427 \dots (- 2.89752 \dots) - 1.75110 \dots = - 9.02614 \dots

f^{^{'}} (u_{4}) = - 4 (- 2.89752 \dots) - 3.28427 \dots = 8.30583 \dots

u_{5} = - 1.81080 \dots

Δ u_{5} = ∣ - 1.81080 \dots - (- 2.89752 \dots) ∣ = 1.08672 \dots

Δ u_{5} = 1.08672 \dots

u_{6} = - 1.21419 \dots : Δ u_{6} = 0.59660 \dots

f (u_{5}) = - 2 (- 1.81080 \dots)^{2} - 3.28427 \dots (- 1.81080 \dots) - 1.75110 \dots = - 2.36193 \dots

f^{^{'}} (u_{5}) = - 4 (- 1.81080 \dots) - 3.28427 \dots = 3.95893 \dots

u_{6} = - 1.21419 \dots

Δ u_{6} = ∣ - 1.21419 \dots - (- 1.81080 \dots) ∣ = 0.59660 \dots

Δ u_{6} = 0.59660 \dots

u_{7} = - 0.76148 \dots : Δ u_{7} = 0.45270 \dots

f (u_{6}) = - 2 (- 1.21419 \dots)^{2} - 3.28427 \dots (- 1.21419 \dots) - 1.75110 \dots = - 0.71188 \dots

f^{^{'}} (u_{6}) = - 4 (- 1.21419 \dots) - 3.28427 \dots = 1.57249 \dots

u_{7} = - 0.76148 \dots

Δ u_{7} = ∣ - 0.76148 \dots - (- 1.21419 \dots) ∣ = 0.45270 \dots

Δ u_{7} = 0.45270 \dots

u_{8} = - 2.48127 \dots : Δ u_{8} = 1.71979 \dots

f (u_{7}) = - 2 (- 0.76148 \dots)^{2} - 3.28427 \dots (- 0.76148 \dots) - 1.75110 \dots = - 0.40989 \dots

f^{^{'}} (u_{7}) = - 4 (- 0.76148 \dots) - 3.28427 \dots = - 0.23833 \dots

u_{8} = - 2.48127 \dots

Δ u_{8} = ∣ - 2.48127 \dots - (- 0.76148 \dots) ∣ = 1.71979 \dots

Δ u_{8} = 1.71979 \dots

u_{9} = - 1.59051 \dots : Δ u_{9} = 0.89075 \dots

f (u_{8}) = - 2 (- 2.48127 \dots)^{2} - 3.28427 \dots (- 2.48127 \dots) - 1.75110 \dots = - 5.91535 \dots

f^{^{'}} (u_{8}) = - 4 (- 2.48127 \dots) - 3.28427 \dots = 6.64082 \dots

u_{9} = - 1.59051 \dots

Δ u_{9} = ∣ - 1.59051 \dots - (- 2.48127 \dots) ∣ = 0.89075 \dots

Δ u_{9} = 0.89075 \dots

u_{10} = - 1.07492 \dots : Δ u_{10} = 0.51559 \dots

f (u_{9}) = - 2 (- 1.59051 \dots)^{2} - 3.28427 \dots (- 1.59051 \dots) - 1.75110 \dots = - 1.58689 \dots

f^{^{'}} (u_{9}) = - 4 (- 1.59051 \dots) - 3.28427 \dots = 3.07779 \dots

u_{10} = - 1.07492 \dots

Δ u_{10} = ∣ - 1.07492 \dots - (- 1.59051 \dots) ∣ = 0.51559 \dots

Δ u_{10} = 0.51559 \dots

u_{11} = - 0.55132 \dots : Δ u_{11} = 0.52360 \dots

f (u_{10}) = - 2 (- 1.07492 \dots)^{2} - 3.28427 \dots (- 1.07492 \dots) - 1.75110 \dots = - 0.53167 \dots

f^{^{'}} (u_{10}) = - 4 (- 1.07492 \dots) - 3.28427 \dots = 1.01541 \dots

u_{11} = - 0.55132 \dots

Δ u_{11} = ∣ - 0.55132 \dots - (- 1.07492 \dots) ∣ = 0.52360 \dots

Δ u_{11} = 0.52360 \dots

u_{12} = - 1.05950 \dots : Δ u_{12} = 0.50817 \dots

f (u_{11}) = - 2 (- 0.55132 \dots)^{2} - 3.28427 \dots (- 0.55132 \dots) - 1.75110 \dots = - 0.54831 \dots

f^{^{'}} (u_{11}) = - 4 (- 0.55132 \dots) - 3.28427 \dots = - 1.07898 \dots

u_{12} = - 1.05950 \dots

Δ u_{12} = ∣ - 1.05950 \dots - (- 0.55132 \dots) ∣ = 0.50817 \dots

Δ u_{12} = 0.50817 \dots

u_{13} = - 0.51794 \dots : Δ u_{13} = 0.54155 \dots

f (u_{12}) = - 2 (- 1.05950 \dots)^{2} - 3.28427 \dots (- 1.05950 \dots) - 1.75110 \dots = - 0.51648 \dots

f^{^{'}} (u_{12}) = - 4 (- 1.05950 \dots) - 3.28427 \dots = 0.95372 \dots

u_{13} = - 0.51794 \dots

Δ u_{13} = ∣ - 0.51794 \dots - (- 1.05950 \dots) ∣ = 0.54155 \dots

Δ u_{13} = 0.54155 \dots

解を見つけられない

解は

u \approx 1.14213 \dots

代用を戻す

u = cos (x)

cos (x) \approx 1.14213 \dots

cos (x) \approx 1.14213 \dots

cos (x) = 1.14213 \dots :

解なし

cos (x) = 1.14213 \dots

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

解なし

1 + \frac{1}{sin ^{2} ( x )} - 2 cos (x) = 0 :

解なし

1 + \frac{1}{sin ^{2} ( x )} - 2 cos (x) = 0

簡素化

1 + \frac{1}{sin ^{2} ( x )} - 2 cos (x) : \frac{sin ^{2} ( x ) + 1 - 2 sin ^{2} ( x ) cos ( x )}{sin ^{2} ( x )}

1 + \frac{1}{sin ^{2} ( x )} - 2 cos (x)

元を分数に変換する:

1 = \frac{1 sin ^{2} ( x )}{sin ^{2} ( x )}, 2 cos (x) = \frac{2 cos ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

= \frac{1 \cdot sin ^{2} ( x )}{sin ^{2} ( x )} + \frac{1}{sin ^{2} ( x )} - \frac{2 cos ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot sin ^{2} ( x ) + 1 - 2 cos ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

乗算:

1 \cdot sin^{2} (x) = sin^{2} (x)

= \frac{sin ^{2} ( x ) + 1 - 2 sin ^{2} ( x ) cos ( x )}{sin ^{2} ( x )}

\frac{sin ^{2} ( x ) + 1 - 2 sin ^{2} ( x ) cos ( x )}{sin ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin^{2} (x) + 1 - 2 sin^{2} (x) cos (x) = 0

三角関数の公式を使用して書き換える

1 + sin^{2} (x) - 2 cos (x) sin^{2} (x)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 + 1 - cos^{2} (x) - 2 cos (x) (1 - cos^{2} (x))

簡素化

1 + 1 - cos^{2} (x) - 2 cos (x) (1 - cos^{2} (x)) : 2 cos^{3} (x) - cos^{2} (x) - 2 cos (x) + 2

1 + 1 - cos^{2} (x) - 2 cos (x) (1 - cos^{2} (x))

拡張

- 2 cos (x) (1 - cos^{2} (x)) : - 2 cos (x) + 2 cos^{3} (x)

- 2 cos (x) (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = - 2 cos (x), b = 1, c = cos^{2} (x)

= - 2 cos (x) \cdot 1 - (- 2 cos (x)) cos^{2} (x)

マイナス・プラスの規則を適用する

- (- a) = a

= - 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x)

簡素化

- 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x) : - 2 cos (x) + 2 cos^{3} (x)

- 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x)

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 \cdot cos (x)

数を乗じる:

2 \cdot 1 = 2

= 2 cos (x)

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

2 cos^{2} (x) cos (x)

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

cos^{2} (x) cos (x) = cos^{2 + 1} (x)

= 2 cos^{2 + 1} (x)

数を足す:

2 + 1 = 3

= 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= 1 + 1 - cos^{2} (x) - 2 cos (x) + 2 cos^{3} (x)

数を足す:

1 + 1 = 2

= 2 cos^{3} (x) - cos^{2} (x) - 2 cos (x) + 2

= 2 cos^{3} (x) - cos^{2} (x) - 2 cos (x) + 2

2 - cos^{2} (x) - 2 cos (x) + 2 cos^{3} (x) = 0

置換で解く

2 - cos^{2} (x) - 2 cos (x) + 2 cos^{3} (x) = 0

仮定:

cos (x) = u

2 - u^{2} - 2 u + 2 u^{3} = 0

2 - u^{2} - 2 u + 2 u^{3} = 0 : u \approx - 1.14213 \dots

2 - u^{2} - 2 u + 2 u^{3} = 0

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{3} - u^{2} - 2 u + 2 = 0

ニュートン・ラプソン法を使用して

2 u^{3} - u^{2} - 2 u + 2 = 0

の解を1つ求める:

u \approx - 1.14213 \dots

2 u^{3} - u^{2} - 2 u + 2 = 0

ニュートン・ラプソン概算の定義

f (u) = 2 u^{3} - u^{2} - 2 u + 2

発見する

f^{^{'}} (u) : 6 u^{2} - 2 u - 2

\frac{d}{d u} (2 u^{3} - u^{2} - 2 u + 2)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (2 u^{3}) - \frac{d}{d u} (u^{2}) - \frac{d}{d u} (2 u) + \frac{d}{d u} (2)

\frac{d}{d u} (2 u^{3}) = 6 u^{2}

\frac{d}{d u} (2 u^{3})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{3})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 3 u^{3 - 1}

簡素化

= 6 u^{2}

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

簡素化

= 2 u

\frac{d}{d u} (2 u) = 2

\frac{d}{d u} (2 u)

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d u}{d u}

共通の導関数を適用:

\frac{d u}{d u} = 1

= 2 \cdot 1

簡素化

= 2

\frac{d}{d u} (2) = 0

\frac{d}{d u} (2)

定数の導関数:

\frac{d}{d x} (a) = 0

= 0

= 6 u^{2} - 2 u - 2 + 0

簡素化

= 6 u^{2} - 2 u - 2

仮定:

u_{0} = 1

Δ u_{n + 1} <

になるまで

u_{n + 1}

を計算する

0.000001

u_{1} = 0.5 : Δ u_{1} = 0.5

f (u_{0}) = 2 \cdot 1^{3} - 1^{2} - 2 \cdot 1 + 2 = 1

f^{^{'}} (u_{0}) = 6 \cdot 1^{2} - 2 \cdot 1 - 2 = 2

u_{1} = 0.5

Δ u_{1} = ∣ 0.5 - 1 ∣ = 0.5

Δ u_{1} = 0.5

u_{2} = 1.16666 \dots : Δ u_{2} = 0.66666 \dots

f (u_{1}) = 2 \cdot 0. 5^{3} - 0. 5^{2} - 2 \cdot 0.5 + 2 = 1

f^{^{'}} (u_{1}) = 6 \cdot 0. 5^{2} - 2 \cdot 0.5 - 2 = - 1.5

u_{2} = 1.16666 \dots

Δ u_{2} = ∣ 1.16666 \dots - 0.5 ∣ = 0.66666 \dots

Δ u_{2} = 0.66666 \dots

u_{3} = 0.78019 \dots : Δ u_{3} = 0.38647 \dots

f (u_{2}) = 2 \cdot 1.16666 \dots^{3} - 1.16666 \dots^{2} - 2 \cdot 1.16666 \dots + 2 = 1.48148 \dots

f^{^{'}} (u_{2}) = 6 \cdot 1.16666 \dots^{2} - 2 \cdot 1.16666 \dots - 2 = 3.83333 \dots

u_{3} = 0.78019 \dots

Δ u_{3} = ∣ 0.78019 \dots - 1.16666 \dots ∣ = 0.38647 \dots

Δ u_{3} = 0.38647 \dots

u_{4} = - 7.72232 \dots : Δ u_{4} = 8.50251 \dots

f (u_{3}) = 2 \cdot 0.78019 \dots^{3} - 0.78019 \dots^{2} - 2 \cdot 0.78019 \dots + 2 = 0.78072 \dots

f^{^{'}} (u_{3}) = 6 \cdot 0.78019 \dots^{2} - 2 \cdot 0.78019 \dots - 2 = 0.09182 \dots

u_{4} = - 7.72232 \dots

Δ u_{4} = ∣ - 7.72232 \dots - 0.78019 \dots ∣ = 8.50251 \dots

Δ u_{4} = 8.50251 \dots

u_{5} = - 5.12779 \dots : Δ u_{5} = 2.59452 \dots

f (u_{4}) = 2 (- 7.72232 \dots)^{3} - (- 7.72232 \dots)^{2} - 2 (- 7.72232 \dots) + 2 = - 963.21890 \dots

f^{^{'}} (u_{4}) = 6 (- 7.72232 \dots)^{2} - 2 (- 7.72232 \dots) - 2 = 371.25003 \dots

u_{5} = - 5.12779 \dots

Δ u_{5} = ∣ - 5.12779 \dots - (- 7.72232 \dots) ∣ = 2.59452 \dots

Δ u_{5} = 2.59452 \dots

u_{6} = - 3.41896 \dots : Δ u_{6} = 1.70882 \dots

f (u_{5}) = 2 (- 5.12779 \dots)^{3} - (- 5.12779 \dots)^{2} - 2 (- 5.12779 \dots) + 2 = - 283.70151 \dots

f^{^{'}} (u_{5}) = 6 (- 5.12779 \dots)^{2} - 2 (- 5.12779 \dots) - 2 = 166.02106 \dots

u_{6} = - 3.41896 \dots

Δ u_{6} = ∣ - 3.41896 \dots - (- 5.12779 \dots) ∣ = 1.70882 \dots

Δ u_{6} = 1.70882 \dots

u_{7} = - 2.31481 \dots : Δ u_{7} = 1.10414 \dots

f (u_{6}) = 2 (- 3.41896 \dots)^{3} - (- 3.41896 \dots)^{2} - 2 (- 3.41896 \dots) + 2 = - 82.78202 \dots

f^{^{'}} (u_{6}) = 6 (- 3.41896 \dots)^{2} - 2 (- 3.41896 \dots) - 2 = 74.97378 \dots

u_{7} = - 2.31481 \dots

Δ u_{7} = ∣ - 2.31481 \dots - (- 3.41896 \dots) ∣ = 1.10414 \dots

Δ u_{7} = 1.10414 \dots

u_{8} = - 1.63810 \dots : Δ u_{8} = 0.67671 \dots

f (u_{7}) = 2 (- 2.31481 \dots)^{3} - (- 2.31481 \dots)^{2} - 2 (- 2.31481 \dots) + 2 = - 23.53607 \dots

f^{^{'}} (u_{7}) = 6 (- 2.31481 \dots)^{2} - 2 (- 2.31481 \dots) - 2 = 34.77990 \dots

u_{8} = - 1.63810 \dots

Δ u_{8} = ∣ - 1.63810 \dots - (- 2.31481 \dots) ∣ = 0.67671 \dots

Δ u_{8} = 0.67671 \dots

u_{9} = - 1.28138 \dots : Δ u_{9} = 0.35671 \dots

f (u_{8}) = 2 (- 1.63810 \dots)^{3} - (- 1.63810 \dots)^{2} - 2 (- 1.63810 \dots) + 2 = - 6.19847 \dots

f^{^{'}} (u_{8}) = 6 (- 1.63810 \dots)^{2} - 2 (- 1.63810 \dots) - 2 = 17.37647 \dots

u_{9} = - 1.28138 \dots

Δ u_{9} = ∣ - 1.28138 \dots - (- 1.63810 \dots) ∣ = 0.35671 \dots

Δ u_{9} = 0.35671 \dots

u_{10} = - 1.15779 \dots : Δ u_{10} = 0.12358 \dots

f (u_{9}) = 2 (- 1.28138 \dots)^{3} - (- 1.28138 \dots)^{2} - 2 (- 1.28138 \dots) + 2 = - 1.28712 \dots

f^{^{'}} (u_{9}) = 6 (- 1.28138 \dots)^{2} - 2 (- 1.28138 \dots) - 2 = 10.41447 \dots

u_{10} = - 1.15779 \dots

Δ u_{10} = ∣ - 1.15779 \dots - (- 1.28138 \dots) ∣ = 0.12358 \dots

Δ u_{10} = 0.12358 \dots

u_{11} = - 1.14237 \dots : Δ u_{11} = 0.01542 \dots

f (u_{10}) = 2 (- 1.15779 \dots)^{3} - (- 1.15779 \dots)^{2} - 2 (- 1.15779 \dots) + 2 = - 0.12893 \dots

f^{^{'}} (u_{10}) = 6 (- 1.15779 \dots)^{2} - 2 (- 1.15779 \dots) - 2 = 8.35854 \dots

u_{11} = - 1.14237 \dots

Δ u_{11} = ∣ - 1.14237 \dots - (- 1.15779 \dots) ∣ = 0.01542 \dots

Δ u_{11} = 0.01542 \dots

u_{12} = - 1.14213 \dots : Δ u_{12} = 0.00023 \dots

f (u_{11}) = 2 (- 1.14237 \dots)^{3} - (- 1.14237 \dots)^{2} - 2 (- 1.14237 \dots) + 2 = - 0.00188 \dots

f^{^{'}} (u_{11}) = 6 (- 1.14237 \dots)^{2} - 2 (- 1.14237 \dots) - 2 = 8.11480 \dots

u_{12} = - 1.14213 \dots

Δ u_{12} = ∣ - 1.14213 \dots - (- 1.14237 \dots) ∣ = 0.00023 \dots

Δ u_{12} = 0.00023 \dots

u_{13} = - 1.14213 \dots : Δ u_{13} = 5.21652 E - 8

f (u_{12}) = 2 (- 1.14213 \dots)^{3} - (- 1.14213 \dots)^{2} - 2 (- 1.14213 \dots) + 2 = - 4.2312 E - 7

f^{^{'}} (u_{12}) = 6 (- 1.14213 \dots)^{2} - 2 (- 1.14213 \dots) - 2 = 8.11116 \dots

u_{13} = - 1.14213 \dots

Δ u_{13} = ∣ - 1.14213 \dots - (- 1.14213 \dots) ∣ = 5.21652 E - 8

Δ u_{13} = 5.21652 E - 8

u \approx - 1.14213 \dots

長除法を適用する:

\frac{2 u ^{3} - u ^{2} - 2 u + 2}{u + 1.14213 \dots} = 2 u^{2} - 3.28427 \dots u + 1.75110 \dots

2 u^{2} - 3.28427 \dots u + 1.75110 \dots \approx 0

ニュートン・ラプソン法を使用して

2 u^{2} - 3.28427 \dots u + 1.75110 \dots = 0

の解を1つ求める:

以下の解はない:

u \in R

2 u^{2} - 3.28427 \dots u + 1.75110 \dots = 0

ニュートン・ラプソン概算の定義

f (u) = 2 u^{2} - 3.28427 \dots u + 1.75110 \dots

発見する

f^{^{'}} (u) : 4 u - 3.28427 \dots

\frac{d}{d u} (2 u^{2} - 3.28427 \dots u + 1.75110 \dots)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (3.28427 \dots u) + \frac{d}{d u} (1.75110 \dots)

\frac{d}{d u} (2 u^{2}) = 4 u

\frac{d}{d u} (2 u^{2})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{2})

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 2 u^{2 - 1}

簡素化

= 4 u

\frac{d}{d u} (3.28427 \dots u) = 3.28427 \dots

\frac{d}{d u} (3.28427 \dots u)

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 3.28427 \dots \frac{d u}{d u}

共通の導関数を適用:

\frac{d u}{d u} = 1

= 3.28427 \dots \cdot 1

簡素化

= 3.28427 \dots

\frac{d}{d u} (1.75110 \dots) = 0

\frac{d}{d u} (1.75110 \dots)

定数の導関数:

\frac{d}{d x} (a) = 0

= 0

= 4 u - 3.28427 \dots + 0

簡素化

= 4 u - 3.28427 \dots

仮定:

u_{0} = 1

Δ u_{n + 1} <

になるまで

u_{n + 1}

を計算する

0.000001

u_{1} = 0.34775 \dots : Δ u_{1} = 0.65224 \dots

f (u_{0}) = 2 \cdot 1^{2} - 3.28427 \dots \cdot 1 + 1.75110 \dots = 0.46682 \dots

f^{^{'}} (u_{0}) = 4 \cdot 1 - 3.28427 \dots = 0.71572 \dots

u_{1} = 0.34775 \dots

Δ u_{1} = ∣ 0.34775 \dots - 1 ∣ = 0.65224 \dots

Δ u_{1} = 0.65224 \dots

u_{2} = 0.79716 \dots : Δ u_{2} = 0.44940 \dots

f (u_{1}) = 2 \cdot 0.34775 \dots^{2} - 3.28427 \dots \cdot 0.34775 \dots + 1.75110 \dots = 0.85083 \dots

f^{^{'}} (u_{1}) = 4 \cdot 0.34775 \dots - 3.28427 \dots = - 1.89323 \dots

u_{2} = 0.79716 \dots

Δ u_{2} = ∣ 0.79716 \dots - 0.34775 \dots ∣ = 0.44940 \dots

Δ u_{2} = 0.44940 \dots

u_{3} = 5.02192 \dots : Δ u_{3} = 4.22476 \dots

f (u_{2}) = 2 \cdot 0.79716 \dots^{2} - 3.28427 \dots \cdot 0.79716 \dots + 1.75110 \dots = 0.40393 \dots

f^{^{'}} (u_{2}) = 4 \cdot 0.79716 \dots - 3.28427 \dots = - 0.09561 \dots

u_{3} = 5.02192 \dots

Δ u_{3} = ∣ 5.02192 \dots - 0.79716 \dots ∣ = 4.22476 \dots

Δ u_{3} = 4.22476 \dots

u_{4} = 2.89752 \dots : Δ u_{4} = 2.12439 \dots

f (u_{3}) = 2 \cdot 5.02192 \dots^{2} - 3.28427 \dots \cdot 5.02192 \dots + 1.75110 \dots = 35.69719 \dots

f^{^{'}} (u_{3}) = 4 \cdot 5.02192 \dots - 3.28427 \dots = 16.80342 \dots

u_{4} = 2.89752 \dots

Δ u_{4} = ∣ 2.89752 \dots - 5.02192 \dots ∣ = 2.12439 \dots

Δ u_{4} = 2.12439 \dots

u_{5} = 1.81080 \dots : Δ u_{5} = 1.08672 \dots

f (u_{4}) = 2 \cdot 2.89752 \dots^{2} - 3.28427 \dots \cdot 2.89752 \dots + 1.75110 \dots = 9.02614 \dots

f^{^{'}} (u_{4}) = 4 \cdot 2.89752 \dots - 3.28427 \dots = 8.30583 \dots

u_{5} = 1.81080 \dots

Δ u_{5} = ∣ 1.81080 \dots - 2.89752 \dots ∣ = 1.08672 \dots

Δ u_{5} = 1.08672 \dots

u_{6} = 1.21419 \dots : Δ u_{6} = 0.59660 \dots

f (u_{5}) = 2 \cdot 1.81080 \dots^{2} - 3.28427 \dots \cdot 1.81080 \dots + 1.75110 \dots = 2.36193 \dots

f^{^{'}} (u_{5}) = 4 \cdot 1.81080 \dots - 3.28427 \dots = 3.95893 \dots

u_{6} = 1.21419 \dots

Δ u_{6} = ∣ 1.21419 \dots - 1.81080 \dots ∣ = 0.59660 \dots

Δ u_{6} = 0.59660 \dots

u_{7} = 0.76148 \dots : Δ u_{7} = 0.45270 \dots

f (u_{6}) = 2 \cdot 1.21419 \dots^{2} - 3.28427 \dots \cdot 1.21419 \dots + 1.75110 \dots = 0.71188 \dots

f^{^{'}} (u_{6}) = 4 \cdot 1.21419 \dots - 3.28427 \dots = 1.57249 \dots

u_{7} = 0.76148 \dots

Δ u_{7} = ∣ 0.76148 \dots - 1.21419 \dots ∣ = 0.45270 \dots

Δ u_{7} = 0.45270 \dots

u_{8} = 2.48127 \dots : Δ u_{8} = 1.71979 \dots

f (u_{7}) = 2 \cdot 0.76148 \dots^{2} - 3.28427 \dots \cdot 0.76148 \dots + 1.75110 \dots = 0.40989 \dots

f^{^{'}} (u_{7}) = 4 \cdot 0.76148 \dots - 3.28427 \dots = - 0.23833 \dots

u_{8} = 2.48127 \dots

Δ u_{8} = ∣ 2.48127 \dots - 0.76148 \dots ∣ = 1.71979 \dots

Δ u_{8} = 1.71979 \dots

u_{9} = 1.59051 \dots : Δ u_{9} = 0.89075 \dots

f (u_{8}) = 2 \cdot 2.48127 \dots^{2} - 3.28427 \dots \cdot 2.48127 \dots + 1.75110 \dots = 5.91535 \dots

f^{^{'}} (u_{8}) = 4 \cdot 2.48127 \dots - 3.28427 \dots = 6.64082 \dots

u_{9} = 1.59051 \dots

Δ u_{9} = ∣ 1.59051 \dots - 2.48127 \dots ∣ = 0.89075 \dots

Δ u_{9} = 0.89075 \dots

u_{10} = 1.07492 \dots : Δ u_{10} = 0.51559 \dots

f (u_{9}) = 2 \cdot 1.59051 \dots^{2} - 3.28427 \dots \cdot 1.59051 \dots + 1.75110 \dots = 1.58689 \dots

f^{^{'}} (u_{9}) = 4 \cdot 1.59051 \dots - 3.28427 \dots = 3.07779 \dots

u_{10} = 1.07492 \dots

Δ u_{10} = ∣ 1.07492 \dots - 1.59051 \dots ∣ = 0.51559 \dots

Δ u_{10} = 0.51559 \dots

u_{11} = 0.55132 \dots : Δ u_{11} = 0.52360 \dots

f (u_{10}) = 2 \cdot 1.07492 \dots^{2} - 3.28427 \dots \cdot 1.07492 \dots + 1.75110 \dots = 0.53167 \dots

f^{^{'}} (u_{10}) = 4 \cdot 1.07492 \dots - 3.28427 \dots = 1.01541 \dots

u_{11} = 0.55132 \dots

Δ u_{11} = ∣ 0.55132 \dots - 1.07492 \dots ∣ = 0.52360 \dots

Δ u_{11} = 0.52360 \dots

u_{12} = 1.05950 \dots : Δ u_{12} = 0.50817 \dots

f (u_{11}) = 2 \cdot 0.55132 \dots^{2} - 3.28427 \dots \cdot 0.55132 \dots + 1.75110 \dots = 0.54831 \dots

f^{^{'}} (u_{11}) = 4 \cdot 0.55132 \dots - 3.28427 \dots = - 1.07898 \dots

u_{12} = 1.05950 \dots

Δ u_{12} = ∣ 1.05950 \dots - 0.55132 \dots ∣ = 0.50817 \dots

Δ u_{12} = 0.50817 \dots

u_{13} = 0.51794 \dots : Δ u_{13} = 0.54155 \dots

f (u_{12}) = 2 \cdot 1.05950 \dots^{2} - 3.28427 \dots \cdot 1.05950 \dots + 1.75110 \dots = 0.51648 \dots

f^{^{'}} (u_{12}) = 4 \cdot 1.05950 \dots - 3.28427 \dots = 0.95372 \dots

u_{13} = 0.51794 \dots

Δ u_{13} = ∣ 0.51794 \dots - 1.05950 \dots ∣ = 0.54155 \dots

Δ u_{13} = 0.54155 \dots

解を見つけられない

解は

u \approx - 1.14213 \dots

代用を戻す

u = cos (x)

cos (x) \approx - 1.14213 \dots

cos (x) \approx - 1.14213 \dots

cos (x) = - 1.14213 \dots :

解なし

cos (x) = - 1.14213 \dots

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

解なし

すべての解を組み合わせる

解なし

元のequationに当てはめて解を検算する

csc^{2} (x) + 2 cos (x) + 1 = 0

に当てはめて解を確認する

equationに一致しないものを削除する。

以下の解はない : x \in R

グラフ

人気の例

2cos^2(2x)+3cos(2x)-2=0

2 cos^{2} (2 x) + 3 cos (2 x) - 2 = 0

tan(t)=-12/5 ,(3pi)/2 <t<2pi

tan (t) = - \frac{12}{5}, \frac{3 π}{2} < t < 2 π

3sin(x)+5cos(x)=4

3 sin (x) + 5 cos (x) = 4

- 2 = sec (x)

2sin^2(x)+4sin(x)=2cos^2(x)+1

2 sin^{2} (x) + 4 sin (x) = 2 cos^{2} (x) + 1