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人気のある三角関数 >

2sin^2(x/2)-cos(x/2)=0

解

2 sin^{2} (\frac{x}{2}) - cos (\frac{x}{2}) = 0

解

x = 2 \cdot 0.67488 \dots + 4 πn, x = 4 π - 2 \cdot 0.67488 \dots + 4 πn

+1

度

x = 77.33656 \dots^{\circ} + 72 0^{\circ} n, x = 642.66343 \dots^{\circ} + 72 0^{\circ} n

解答ステップ

2 sin^{2} (\frac{x}{2}) - cos (\frac{x}{2}) = 0

三角関数の公式を使用して書き換える

- cos (\frac{x}{2}) + 2 sin^{2} (\frac{x}{2})

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)

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

- cos (\frac{x}{2}) + (1 - cos^{2} (\frac{x}{2})) \cdot 2 = 0

置換で解く

- cos (\frac{x}{2}) + (1 - cos^{2} (\frac{x}{2})) \cdot 2 = 0

仮定:

cos (\frac{x}{2}) = u

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

- u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

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

拡張

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

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

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

拡張

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

2 (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

= 2 - 2 u^{2}

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 2, b = - 1, c = 2

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

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

(- 1)^{2} - 4 (- 2) \cdot 2 = 17

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

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

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

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

数を乗じる:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

数を足す:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 2 )} : - \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 ( - 2 )}

括弧を削除する:

(- a) = - a, - (- a) = a

= \frac{1 + 17}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{1 + 17}{- 4}

分数の規則を適用する:

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

= - \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 2 )} : \frac{17 - 1}{4}

\frac{- ( - 1 ) - 17}{2 ( - 2 )}

括弧を削除する:

(- a) = - a, - (- a) = a

= \frac{1 - 17}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{1 - 17}{- 4}

分数の規則を適用する:

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

1 - 17 = - (17 - 1)

= \frac{17 - 1}{4}

二次equationの解:

u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

代用を戻す

u = cos (\frac{x}{2})

cos (\frac{x}{2}) = - \frac{1 + 17}{4}, cos (\frac{x}{2}) = \frac{17 - 1}{4}

cos (\frac{x}{2}) = - \frac{1 + 17}{4}, cos (\frac{x}{2}) = \frac{17 - 1}{4}

cos (\frac{x}{2}) = - \frac{1 + 17}{4} :

解なし

cos (\frac{x}{2}) = - \frac{1 + 17}{4}

- 1 \leq cos (x) \leq 1

解なし

cos (\frac{x}{2}) = \frac{17 - 1}{4} : x = 2 arccos (\frac{17 - 1}{4}) + 4 πn, x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

cos (\frac{x}{2}) = \frac{17 - 1}{4}

三角関数の逆数プロパティを適用する

cos (\frac{x}{2}) = \frac{17 - 1}{4}

以下の一般解

cos (\frac{x}{2}) = \frac{17 - 1}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn, \frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn, \frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

解く

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn : x = 2 arccos (\frac{17 - 1}{4}) + 4 πn

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn

以下で両辺を乗じる:

2

\frac{x}{2} = arccos (\frac{17 - 1}{4}) + 2 πn

以下で両辺を乗じる:

2

\frac{2 x}{2} = 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

簡素化

\frac{2 x}{2} = 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn : 2 arccos (\frac{17 - 1}{4}) + 4 πn

2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn

解く

\frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn : x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

\frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

以下で両辺を乗じる:

2

\frac{x}{2} = 2 π - arccos (\frac{17 - 1}{4}) + 2 πn

以下で両辺を乗じる:

2

\frac{2 x}{2} = 2 \cdot 2 π - 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

簡素化

\frac{2 x}{2} = 2 \cdot 2 π - 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

2 \cdot 2 π - 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn : 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

2 \cdot 2 π - 2 arccos (\frac{17 - 1}{4}) + 2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn, x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

すべての解を組み合わせる

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn, x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

10進法形式で解を証明する

x = 2 \cdot 0.67488 \dots + 4 πn, x = 4 π - 2 \cdot 0.67488 \dots + 4 πn

グラフ

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