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人気のある三角関数 >

solvefor x,sin(x)=a-2qcos(2x)

解

解く

x, sin (x) = a - 2 \cdot q \cdot cos (2 x)

解

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

解答ステップ

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

両辺から

a - 2 q cos (2 x)

を引く

sin (x) - a + 2 q cos (2 x) = 0

三角関数の公式を使用して書き換える

sin (x) - a + 2 cos (2 x) q

2倍角の公式を使用:

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

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

sin (x) - a + (1 - 2 sin^{2} (x)) \cdot 2 q = 0

置換で解く

sin (x) - a + (1 - 2 sin^{2} (x)) \cdot 2 q = 0

仮定:

sin (x) = u

u - a + (1 - 2 u^{2}) \cdot 2 q = 0

u - a + (1 - 2 u^{2}) \cdot 2 q = 0 : u = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, u = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

u - a + (1 - 2 u^{2}) \cdot 2 q = 0

拡張

u - a + (1 - 2 u^{2}) \cdot 2 q : u - a + 2 q - 4 q u^{2}

u - a + (1 - 2 u^{2}) \cdot 2 q

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

拡張

2 q (1 - 2 u^{2}) : 2 q - 4 q u^{2}

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

分配法則を適用する:

a (b - c) = ab - a c

a = 2 q, b = 1, c = 2 u^{2}

= 2 q \cdot 1 - 2 q \cdot 2 u^{2}

= 2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2}

簡素化

2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2} : 2 q - 4 q u^{2}

2 \cdot 1 \cdot q - 2 \cdot 2 q u^{2}

数を乗じる:

2 \cdot 1 = 2

= 2 q - 2 \cdot 2 q u^{2}

数を乗じる:

2 \cdot 2 = 4

= 2 q - 4 q u^{2}

= 2 q - 4 q u^{2}

= u - a + 2 q - 4 q u^{2}

u - a + 2 q - 4 q u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

- 4 q u^{2} + u - a + 2 q = 0

解くとthe二次式

- 4 q u^{2} + u - a + 2 q = 0

二次Equationの公式:

次の場合:

a = - 4 q, b = 1, c = - a + 2 q

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 4 q ) ( - a + 2 q )}{2 ( - 4 q )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 4 q ) ( - a + 2 q )}{2 ( - 4 q )}

簡素化

1^{2} - 4 (- 4 q) (- a + 2 q) : 1 + 16 q (- a + 2 q)

1^{2} - 4 (- 4 q) (- a + 2 q)

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 4 q) (2 q - a)

規則を適用

- (- a) = a

= 1 + 4 \cdot 4 q (- a + 2 q)

数を乗じる:

4 \cdot 4 = 16

= 1 + 16 q (2 q - a)

u_{1, 2} = \frac{- 1 \pm 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}; q \neq = 0

解を分離する

u_{1} = \frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}, u_{2} = \frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

u = \frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )} : - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

\frac{- 1 + 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

括弧を削除する:

(- a) = - a

= \frac{- 1 + 1 + 16 q ( - a + 2 q )}{- 2 \cdot 4 q}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 1 + 16 q ( 2 q - a ) + 1}{- 8 q}

分数の規則を適用する:

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

= - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

u = \frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )} : - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

\frac{- 1 - 1 + 16 q ( - a + 2 q )}{2 ( - 4 q )}

括弧を削除する:

(- a) = - a

= \frac{- 1 - 1 + 16 q ( - a + 2 q )}{- 2 \cdot 4 q}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 1 - 16 q ( 2 q - a ) + 1}{- 8 q}

分数の規則を適用する:

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

= - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

二次equationの解:

u = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, u = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

代用を戻す

u = sin (x)

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q} : x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

以下の一般解

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q} : x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

以下の一般解

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

すべての解を組み合わせる

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

グラフ

人気の例

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