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人気のある三角関数 >

sin(3x)+sin(x)=2cos^2(x)

解

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

解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{2} + 2 πn

+1

度

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

2 cos^{2} (x)

を引く

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

三角関数の公式を使用して書き換える

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

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

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

sin (3 x)

三角関数の公式を使用して書き換える

sin (3 x)

書き換え

= sin (2 x + x)

角の和の公式を使用する:

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

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

2倍角の公式を使用:

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

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

簡素化

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

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

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

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

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

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

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

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

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

2倍角の公式を使用:

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

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

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

拡張

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

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

簡素化

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

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

1 \cdot sin (x) = sin (x)

1 sin (x)

乗算:

1 \cdot sin (x) = sin (x)

= sin (x)

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

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

指数の規則を適用する:

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

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

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

数を足す:

2 + 1 = 3

= 2 sin^{3} (x)

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

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

簡素化

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

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

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

2 \cdot 1 sin (x)

数を乗じる:

2 \cdot 1 = 2

= 2 sin (x)

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

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

指数の規則を適用する:

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

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

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

数を足す:

2 + 1 = 3

= 2 sin^{3} (x)

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

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

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

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

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

類似した元を足す:

sin (x) + 2 sin (x) = 3 sin (x)

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

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

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

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

マイナス・プラスの規則を適用する

- (- a) = a

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

数を乗じる:

2 \cdot 1 = 2

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

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

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

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

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

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

標準的な形式で書く

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

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

因数

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

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

共通項をくくり出す

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

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

4

を書き換え

2 \cdot 2

4

を書き換え

2 \cdot 2

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

共通項をくくり出す

- 2

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

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

因数

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

2 u^{3} - u^{2} - 2 u + 1

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

- 1

を

- 2 u + 1 : - (2 u - 1)

からくくり出す

- 2 u + 1

共通項をくくり出す

- 1

= - (2 u - 1)

u^{2}

を

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

からくくり出す

2 u^{3} - u^{2}

指数の規則を適用する:

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

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

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

共通項をくくり出す

u^{2}

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

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

共通項をくくり出す

2 u - 1

= (2 u - 1) (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)

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

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

- 2 (2 u - 1) (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

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

解く

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

u = \frac{1}{2}

u = \frac{1}{2}

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

代用を戻す

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

以下の一般解

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

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

以下の一般解

sin (x) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

以下の一般解

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

すべての解を組み合わせる

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{2} + 2 πn

グラフ

人気の例

tan (θ) = \frac{5}{9}

2sin(x)+3cos(x)=0

2 sin (x) + 3 cos (x) = 0

tan(x)+cot(x)=2sqrt(2)

tan (x) + cot (x) = 22

sin (30 t) = - 0.6

2cos(x)+2sqrt(2)=3sec(x)

2 cos (x) + 22 = 3 sec (x)