ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある >

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

解

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

解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

+1

度

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

2 u^{3} - u = 0

2 u^{3} - u = 0 : u = 0, u = - \frac{2}{2}, u = \frac{2}{2}

2 u^{3} - u = 0

因数

2 u^{3} - u : u (2 u + 1) (2 u - 1)

2 u^{3} - u

共通項をくくり出す

u : u (2 u^{2} - 1)

2 u^{3} - u

指数の規則を適用する:

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

u^{3} = u^{2} u

= 2 u^{2} u - u

共通項をくくり出す

u

= u (2 u^{2} - 1)

= u (2 u^{2} - 1)

因数

2 u^{2} - 1 : (2 u + 1) (2 u - 1)

2 u^{2} - 1

2 u^{2} - 1

を書き換え

(2 u)^{2} - 1^{2}

2 u^{2} - 1

累乗根の規則を適用する:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} u^{2} - 1

1

を書き換え

1^{2}

= (2)^{2} u^{2} - 1^{2}

指数の規則を適用する:

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

(2)^{2} u^{2} = (2 u)^{2}

= (2 u)^{2} - 1^{2}

= (2 u)^{2} - 1^{2}

2乗の差の公式を適用する:

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

(2 u)^{2} - 1^{2} = (2 u + 1) (2 u - 1)

= (2 u + 1) (2 u - 1)

= u (2 u + 1) (2 u - 1)

u (2 u + 1) (2 u - 1) = 0

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

u = 0 or 2 u + 1 = 0 or 2 u - 1 = 0

解く

2 u + 1 = 0 : u = - \frac{2}{2}

2 u + 1 = 0

1

を右側に移動します

2 u + 1 = 0

両辺から

1

を引く

2 u + 1 - 1 = 0 - 1

簡素化

2 u = - 1

2 u = - 1

以下で両辺を割る

2

2 u = - 1

以下で両辺を割る

2

\frac{2 u}{2} = \frac{- 1}{2}

簡素化

\frac{2 u}{2} = \frac{- 1}{2}

簡素化

\frac{2 u}{2} : u

\frac{2 u}{2}

共通因数を約分する:

2

= u

簡素化

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

分数の規則を適用する:

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

= - \frac{1}{2}

有理化する

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

共役で乗じる

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

累乗根の規則を適用する:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

u = - \frac{2}{2}

解く

2 u - 1 = 0 : u = \frac{2}{2}

2 u - 1 = 0

1

を右側に移動します

2 u - 1 = 0

両辺に

1

を足す

2 u - 1 + 1 = 0 + 1

簡素化

2 u = 1

2 u = 1

以下で両辺を割る

2

2 u = 1

以下で両辺を割る

2

\frac{2 u}{2} = \frac{1}{2}

簡素化

\frac{2 u}{2} = \frac{1}{2}

簡素化

\frac{2 u}{2} : u

\frac{2 u}{2}

共通因数を約分する:

2

= u

簡素化

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

共役で乗じる

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

累乗根の規則を適用する:

a a = a

22 = 2

= 2

= \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

u = \frac{2}{2}

解答は

u = 0, u = - \frac{2}{2}, u = \frac{2}{2}

代用を戻す

u = cos (x)

cos (x) = 0, cos (x) = - \frac{2}{2}, cos (x) = \frac{2}{2}

cos (x) = 0, cos (x) = - \frac{2}{2}, cos (x) = \frac{2}{2}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

以下の一般解

cos (x) = 0

cos (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = - \frac{2}{2} : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

cos (x) = - \frac{2}{2}

以下の一般解

cos (x) = - \frac{2}{2}

cos (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

cos (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

cos (x) = \frac{2}{2}

以下の一般解

cos (x) = \frac{2}{2}

cos (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

すべての解を組み合わせる

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

グラフ

人気の例

-2cos(2x)=sqrt(3)

- 2 cos (2 x) = 3

(tan(x)-1)(2sin(x)-sqrt(3))=0

(tan (x) - 1) (2 sin (x) - 3) = 0

tan(θ)(1+tan^2(θ))=0

tan (θ) (1 + tan^{2} (θ)) = 0

tan (x) = 9

sin(2x)=0,0<= x<= 2pi

sin (2 x) = 0, 0 \leq x \leq 2 π