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人気のある三角関数 >

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

解

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

解

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

+1

度

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

解答ステップ

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

両辺から

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

を引く

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

三角関数の公式を使用して書き換える

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

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

簡素化

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

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

置換で解く

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

仮定:

cos (x) = u

- 4 u^{2} + 4 u^{3} = 0

- 4 u^{2} + 4 u^{3} = 0 : u = 0, u = 1

- 4 u^{2} + 4 u^{3} = 0

因数

- 4 u^{2} + 4 u^{3} : 4 u^{2} (u - 1)

- 4 u^{2} + 4 u^{3}

指数の規則を適用する:

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

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

= 4 u u^{2} - 4 u^{2}

共通項をくくり出す

4 u^{2}

= 4 u^{2} (u - 1)

4 u^{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 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

両辺に

1

を足す

u - 1 + 1 = 0 + 1

簡素化

u = 1

u = 1

解答は

u = 0, u = 1

代用を戻す

u = cos (x)

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

cos (x) = 0, 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 = 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 = 2 πn

グラフ

人気の例

2 sin (2 θ) = 0

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

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

sin(8x+2)=cos(6x-10)

sin (8 x + 2) = cos (6 x - 10)

cos(2θ-pi/2)=-1,0<= θ<= 2pi

cos (2 θ - \frac{π}{2}) = - 1, 0 \leq θ \leq 2 π

sin^2(2x)=2sin^2(x)

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