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人気のある三角関数 >

2sin(x)cos^2(x)-1+2cos^2(x)-sin(x)=0

解

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

解

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

+1

度

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

解答ステップ

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

三角関数の公式を使用して書き換える

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

簡素化

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

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

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

数を乗じる:

2 \cdot 1 = 2

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

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

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

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

2 \cdot 1 \cdot 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)

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

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

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

数を足す/引く:

- 1 + 2 = 1

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

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

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

標準的な形式で書く

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

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

因数

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

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

共通項をくくり出す

- 1

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

因数

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

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

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

- 1

を

- u - 1 : - (u + 1)

からくくり出す

- u - 1

共通項をくくり出す

- 1

= - (u + 1)

2 u^{2}

を

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

からくくり出す

2 u^{3} + 2 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} + 2 u^{2}

共通項をくくり出す

2 u^{2}

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

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

共通項をくくり出す

u + 1

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

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

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

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

代用を戻す

u = sin (x)

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

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

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) = - \frac{2}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

以下の一般解

sin (x) = - \frac{2}{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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

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

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

以下の一般解

sin (x) = \frac{2}{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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

すべての解を組み合わせる

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

グラフ

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