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人気のある三角関数 >

3cos^2(x)+5sin(x)-4=0

解

3 cos^{2} (x) + 5 sin (x) - 4 = 0

解

x = 0.23455 \dots + 2 πn, x = π - 0.23455 \dots + 2 πn

+1

度

x = 13.43888 \dots^{\circ} + 36 0^{\circ} n, x = 166.56111 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

3 cos^{2} (x) + 5 sin (x) - 4 = 0

三角関数の公式を使用して書き換える

- 4 + 3 cos^{2} (x) + 5 sin (x)

ピタゴラスの公式を使用する:

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

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

= - 4 + 3 (1 - sin^{2} (x)) + 5 sin (x)

簡素化

- 4 + 3 (1 - sin^{2} (x)) + 5 sin (x) : 5 sin (x) - 3 sin^{2} (x) - 1

- 4 + 3 (1 - sin^{2} (x)) + 5 sin (x)

拡張

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

3 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

数を乗じる:

3 \cdot 1 = 3

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

= - 4 + 3 - 3 sin^{2} (x) + 5 sin (x)

数を足す/引く:

- 4 + 3 = - 1

= 5 sin (x) - 3 sin^{2} (x) - 1

= 5 sin (x) - 3 sin^{2} (x) - 1

- 1 - 3 sin^{2} (x) + 5 sin (x) = 0

置換で解く

- 1 - 3 sin^{2} (x) + 5 sin (x) = 0

仮定:

sin (x) = u

- 1 - 3 u^{2} + 5 u = 0

- 1 - 3 u^{2} + 5 u = 0 : u = \frac{5 - 13}{6}, u = \frac{5 + 13}{6}

- 1 - 3 u^{2} + 5 u = 0

標準的な形式で書く

a x^{2} + b x + c = 0

- 3 u^{2} + 5 u - 1 = 0

解くとthe二次式

- 3 u^{2} + 5 u - 1 = 0

二次Equationの公式:

次の場合:

a = - 3, b = 5, c = - 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 3 ) ( - 1 )}{2 ( - 3 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 3 ) ( - 1 )}{2 ( - 3 )}

5^{2} - 4 (- 3) (- 1) = 13

5^{2} - 4 (- 3) (- 1)

規則を適用

- (- a) = a

= 5^{2} - 4 \cdot 3 \cdot 1

数を乗じる:

4 \cdot 3 \cdot 1 = 12

= 5^{2} - 12

5^{2} = 25

= 25 - 12

数を引く:

25 - 12 = 13

= 13

u_{1, 2} = \frac{- 5 \pm 13}{2 ( - 3 )}

解を分離する

u_{1} = \frac{- 5 + 13}{2 ( - 3 )}, u_{2} = \frac{- 5 - 13}{2 ( - 3 )}

u = \frac{- 5 + 13}{2 ( - 3 )} : \frac{5 - 13}{6}

\frac{- 5 + 13}{2 ( - 3 )}

括弧を削除する:

(- a) = - a

= \frac{- 5 + 13}{- 2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{- 5 + 13}{- 6}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

- 5 + 13 = - (5 - 13)

= \frac{5 - 13}{6}

u = \frac{- 5 - 13}{2 ( - 3 )} : \frac{5 + 13}{6}

\frac{- 5 - 13}{2 ( - 3 )}

括弧を削除する:

(- a) = - a

= \frac{- 5 - 13}{- 2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{- 5 - 13}{- 6}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

- 5 - 13 = - (5 + 13)

= \frac{5 + 13}{6}

二次equationの解:

u = \frac{5 - 13}{6}, u = \frac{5 + 13}{6}

代用を戻す

u = sin (x)

sin (x) = \frac{5 - 13}{6}, sin (x) = \frac{5 + 13}{6}

sin (x) = \frac{5 - 13}{6}, sin (x) = \frac{5 + 13}{6}

sin (x) = \frac{5 - 13}{6} : x = arcsin (\frac{5 - 13}{6}) + 2 πn, x = π - arcsin (\frac{5 - 13}{6}) + 2 πn

sin (x) = \frac{5 - 13}{6}

三角関数の逆数プロパティを適用する

sin (x) = \frac{5 - 13}{6}

以下の一般解

sin (x) = \frac{5 - 13}{6}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{5 - 13}{6}) + 2 πn, x = π - arcsin (\frac{5 - 13}{6}) + 2 πn

x = arcsin (\frac{5 - 13}{6}) + 2 πn, x = π - arcsin (\frac{5 - 13}{6}) + 2 πn

sin (x) = \frac{5 + 13}{6} :

解なし

sin (x) = \frac{5 + 13}{6}

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

x = arcsin (\frac{5 - 13}{6}) + 2 πn, x = π - arcsin (\frac{5 - 13}{6}) + 2 πn

10進法形式で解を証明する

x = 0.23455 \dots + 2 πn, x = π - 0.23455 \dots + 2 πn

グラフ

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