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人気のある三角関数 >

(2-2sin(x))/a = 2/(sin(x))

解

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

解

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

解答ステップ

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

置換で解く

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

仮定:

sin (x) = u

\frac{2 - 2 u}{a} = \frac{2}{u}

\frac{2 - 2 u}{a} = \frac{2}{u} : u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

\frac{2 - 2 u}{a} = \frac{2}{u}

たすき掛け

\frac{2 - 2 u}{a} = \frac{2}{u}

分数たすき掛けを適用する:

\frac{a}{b} = \frac{c}{d}

ならば,

a \cdot d = b \cdot c

(2 - 2 u) u = a \cdot 2

(2 - 2 u) u = a \cdot 2

解く

(2 - 2 u) u = a \cdot 2 : u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}

(2 - 2 u) u = a \cdot 2

拡張

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

(2 - 2 u) u

= u (2 - 2 u)

分配法則を適用する:

a (b - c) = ab - a c

a = u, b = 2, c = 2 u

= u \cdot 2 - u \cdot 2 u

= 2 u - 2 uu

2 uu = 2 u^{2}

2 uu

指数の規則を適用する:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

数を足す:

1 + 1 = 2

= 2 u^{2}

= 2 u - 2 u^{2}

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

a 2

を左側に移動します

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

両辺から

a 2

を引く

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

簡素化

2 u - 2 u^{2} - a \cdot 2 = 0

2 u - 2 u^{2} - a \cdot 2 = 0

標準的な形式で書く

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

- 2 u^{2} + 2 u - a \cdot 2 = 0

解くとthe二次式

- 2 u^{2} + 2 u - a \cdot 2 = 0

二次Equationの公式:

次の場合:

a = - 2, b = 2, c = - a 2

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

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

簡素化

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

2^{2} - 4 (- 2) (- a \cdot 2)

規則を適用

- (- a) = a

= 2^{2} - 4 \cdot 2 a \cdot 2

数を乗じる:

4 \cdot 2 \cdot 2 = 16

= 2^{2} - 16 a

因数

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

2^{2} - 16 a

書き換え

= 4 \cdot 1 - 4 \cdot 4 a

共通項をくくり出す

4

= 4 (1 - 4 a)

= 4 (1 - 4 a)

累乗根の規則を適用する:

n ab = n a n b,

, 以下を想定

a \geq 0, b \geq 0

= 4 - 4 a + 1

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= 2 - 4 a + 1

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

解を分離する

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

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

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

括弧を削除する:

(- a) = - a

= \frac{- 2 + 2 1 - 4 a}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2 + 2 - 4 a + 1}{- 4}

分数の規則を適用する:

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

= - \frac{- 2 + 2 1 - 4 a}{4}

キャンセル

\frac{- 2 + 2 1 - 4 a}{4} : \frac{- 4 a + 1 - 1}{2}

\frac{- 2 + 2 1 - 4 a}{4}

因数

- 2 + 2 1 - 4 a : 2 (- 1 + 1 - 4 a)

- 2 + 2 1 - 4 a

書き換え

= - 2 \cdot 1 + 2 1 - 4 a

共通項をくくり出す

2

= 2 (- 1 + 1 - 4 a)

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

共通因数を約分する:

2

= \frac{- 1 + - 4 a + 1}{2}

= - \frac{- 4 a + 1 - 1}{2}

= - \frac{- 1 + - 4 a + 1}{2}

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

\frac{- 2 - 2 1 - 4 a}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 2 - 2 1 - 4 a}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2 - 2 - 4 a + 1}{- 4}

分数の規則を適用する:

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

= - \frac{- 2 - 2 1 - 4 a}{4}

キャンセル

\frac{- 2 - 2 1 - 4 a}{4} : - \frac{- 4 a + 1 + 1}{2}

\frac{- 2 - 2 1 - 4 a}{4}

因数

- 2 - 2 1 - 4 a : - 2 (1 + 1 - 4 a)

- 2 - 2 1 - 4 a

書き換え

= - 2 \cdot 1 - 2 1 - 4 a

共通項をくくり出す

2

= - 2 (1 + 1 - 4 a)

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

共通因数を約分する:

2

= - \frac{- 4 a + 1 + 1}{2}

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

規則を適用

- (- a) = a

= \frac{- 4 a + 1 + 1}{2}

二次equationの解:

u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}

u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

代用を戻す

u = sin (x)

sin (x) = - \frac{- 1 + - 4 a + 1}{2}, sin (x) = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

sin (x) = - \frac{- 1 + - 4 a + 1}{2}, sin (x) = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

sin (x) = - \frac{- 1 + - 4 a + 1}{2} : x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

以下の一般解

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

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

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn

sin (x) = \frac{- 4 a + 1 + 1}{2} : x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

sin (x) = \frac{- 4 a + 1 + 1}{2}

三角関数の逆数プロパティを適用する

sin (x) = \frac{- 4 a + 1 + 1}{2}

以下の一般解

sin (x) = \frac{- 4 a + 1 + 1}{2}

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

x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

すべての解を組み合わせる

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

グラフ

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