ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

solvefor x,2cos(x)=2cos(3x)

解

解く

x, 2 cos (x) = 2 cos (3 x)

解

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

+1

度

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 cos (x) = 2 cos (3 x)

両辺から

2 cos (3 x)

を引く

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

三角関数の公式を使用して書き換える

- 2 cos (3 x) + 2 cos (x)

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

三角関数の公式を使用して書き換える

cos (3 x)

書き換え

= cos (2 x + x)

角の和の公式を使用する:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 x) cos (x) - sin (2 x) sin (x)

2倍角の公式を使用:

sin (2 x) = 2 sin (x) cos (x)

= cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

簡素化

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

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

2 sin (x) cos (x) sin (x)

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

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

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

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

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

2倍角の公式を使用:

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

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

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

拡張

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x) : 4 cos^{3} (x) - 3 cos (x)

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

簡素化

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

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

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

1 cos (x)

乗算:

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

= cos (x)

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

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

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

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

2 \cdot 1 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)

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x)

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

類似した元を足す:

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

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

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

簡素化

- 2 (4 cos^{3} (x) - 3 cos (x)) + 2 cos (x) : - 8 cos^{3} (x) + 8 cos (x)

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

拡張

- 2 (4 cos^{3} (x) - 3 cos (x)) : - 8 cos^{3} (x) + 6 cos (x)

- 2 (4 cos^{3} (x) - 3 cos (x))

分配法則を適用する:

a (b - c) = ab - a c

a = - 2, b = 4 cos^{3} (x), c = 3 cos (x)

= - 2 \cdot 4 cos^{3} (x) - (- 2) \cdot 3 cos (x)

マイナス・プラスの規則を適用する

- (- a) = a

= - 2 \cdot 4 cos^{3} (x) + 2 \cdot 3 cos (x)

簡素化

- 2 \cdot 4 cos^{3} (x) + 2 \cdot 3 cos (x) : - 8 cos^{3} (x) + 6 cos (x)

- 2 \cdot 4 cos^{3} (x) + 2 \cdot 3 cos (x)

数を乗じる:

2 \cdot 4 = 8

= - 8 cos^{3} (x) + 2 \cdot 3 cos (x)

数を乗じる:

2 \cdot 3 = 6

= - 8 cos^{3} (x) + 6 cos (x)

= - 8 cos^{3} (x) + 6 cos (x)

= - 8 cos^{3} (x) + 6 cos (x) + 2 cos (x)

類似した元を足す:

6 cos (x) + 2 cos (x) = 8 cos (x)

= - 8 cos^{3} (x) + 8 cos (x)

= - 8 cos^{3} (x) + 8 cos (x)

8 cos (x) - 8 cos^{3} (x) = 0

置換で解く

8 cos (x) - 8 cos^{3} (x) = 0

仮定:

cos (x) = u

8 u - 8 u^{3} = 0

8 u - 8 u^{3} = 0 : u = 0, u = - 1, u = 1

8 u - 8 u^{3} = 0

因数

8 u - 8 u^{3} : - 8 u (u + 1) (u - 1)

8 u - 8 u^{3}

共通項をくくり出す

- 8 u : - 8 u (u^{2} - 1)

- 8 u^{3} + 8 u

指数の規則を適用する:

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

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

= - 8 u^{2} u + 8 u

共通項をくくり出す

- 8 u

= - 8 u (u^{2} - 1)

= - 8 u (u^{2} - 1)

因数

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

u^{2} - 1

1

を書き換え

1^{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)

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

= (u + 1) (u - 1)

= - 8 u (u + 1) (u - 1)

- 8 u (u + 1) (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 u + 1 = 0 or u - 1 = 0

解く

u + 1 = 0 : u = - 1

u + 1 = 0

1

を右側に移動します

u + 1 = 0

両辺から

1

を引く

u + 1 - 1 = 0 - 1

簡素化

u = - 1

u = - 1

解く

u - 1 = 0 : u = 1

u - 1 = 0

1

を右側に移動します

u - 1 = 0

両辺に

1

を足す

u - 1 + 1 = 0 + 1

簡素化

u = 1

u = 1

解答は

u = 0, u = - 1, u = 1

代用を戻す

u = cos (x)

cos (x) = 0, cos (x) = - 1, cos (x) = 1

cos (x) = 0, cos (x) = - 1, cos (x) = 1

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) = - 1 : x = π + 2 πn

cos (x) = - 1

以下の一般解

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

以下の一般解

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 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 = π + 2 πn, x = 2 πn

グラフ

人気の例

sin(x)=1-2sin^2(x)

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

2sin^2(t)-cos(t)-1=0

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

4cos^2(θ)=1,0<= θ<2pi

4 cos^{2} (θ) = 1, 0 \leq θ < 2 π

cos(x/2+pi/3)=(sqrt(2))/2

cos (\frac{x}{2} + \frac{π}{3}) = \frac{2}{2}

tan (θ) = 8.2