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人気のある三角関数 >

-cos(x)+4cos(2x)=0

解

- cos (x) + 4 cos (2 x) = 0

解

x = 0.68824 \dots + 2 πn, x = 2 π - 0.68824 \dots + 2 πn, x = 2.27491 \dots + 2 πn, x = - 2.27491 \dots + 2 πn

+1

度

x = 39.43339 \dots^{\circ} + 36 0^{\circ} n, x = 320.56660 \dots^{\circ} + 36 0^{\circ} n, x = 130.34311 \dots^{\circ} + 36 0^{\circ} n, x = - 130.34311 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

- cos (x) + 4 cos (2 x) = 0

三角関数の公式を使用して書き換える

- cos (x) + 4 cos (2 x)

2倍角の公式を使用:

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

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

- cos (x) + (- 1 + 2 cos^{2} (x)) \cdot 4 = 0

置換で解く

- cos (x) + (- 1 + 2 cos^{2} (x)) \cdot 4 = 0

仮定:

cos (x) = u

- u + (- 1 + 2 u^{2}) \cdot 4 = 0

- u + (- 1 + 2 u^{2}) \cdot 4 = 0 : u = \frac{1 + 129}{16}, u = \frac{1 - 129}{16}

- u + (- 1 + 2 u^{2}) \cdot 4 = 0

拡張

- u + (- 1 + 2 u^{2}) \cdot 4 : - u - 4 + 8 u^{2}

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

= - u + 4 (- 1 + 2 u^{2})

拡張

4 (- 1 + 2 u^{2}) : - 4 + 8 u^{2}

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

分配法則を適用する:

a (b + c) = ab + a c

a = 4, b = - 1, c = 2 u^{2}

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

マイナス・プラスの規則を適用する

+ (- a) = - a

= - 4 \cdot 1 + 4 \cdot 2 u^{2}

簡素化

- 4 \cdot 1 + 4 \cdot 2 u^{2} : - 4 + 8 u^{2}

- 4 \cdot 1 + 4 \cdot 2 u^{2}

数を乗じる:

4 \cdot 1 = 4

= - 4 + 4 \cdot 2 u^{2}

数を乗じる:

4 \cdot 2 = 8

= - 4 + 8 u^{2}

= - 4 + 8 u^{2}

= - u - 4 + 8 u^{2}

- u - 4 + 8 u^{2} = 0

標準的な形式で書く

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

8 u^{2} - u - 4 = 0

解くとthe二次式

8 u^{2} - u - 4 = 0

二次Equationの公式:

次の場合:

a = 8, b = - 1, c = - 4

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 8 ( - 4 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 8 ( - 4 )}{2 \cdot 8}

(- 1)^{2} - 4 \cdot 8 (- 4) = 129

(- 1)^{2} - 4 \cdot 8 (- 4)

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 8 \cdot 4

(- 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 8 \cdot 4 = 128

4 \cdot 8 \cdot 4

数を乗じる:

4 \cdot 8 \cdot 4 = 128

= 128

= 1 + 128

数を足す:

1 + 128 = 129

= 129

u_{1, 2} = \frac{- ( - 1 ) \pm 129}{2 \cdot 8}

解を分離する

u_{1} = \frac{- ( - 1 ) + 129}{2 \cdot 8}, u_{2} = \frac{- ( - 1 ) - 129}{2 \cdot 8}

u = \frac{- ( - 1 ) + 129}{2 \cdot 8} : \frac{1 + 129}{16}

\frac{- ( - 1 ) + 129}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{1 + 129}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{1 + 129}{16}

u = \frac{- ( - 1 ) - 129}{2 \cdot 8} : \frac{1 - 129}{16}

\frac{- ( - 1 ) - 129}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{1 - 129}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{1 - 129}{16}

二次equationの解:

u = \frac{1 + 129}{16}, u = \frac{1 - 129}{16}

代用を戻す

u = cos (x)

cos (x) = \frac{1 + 129}{16}, cos (x) = \frac{1 - 129}{16}

cos (x) = \frac{1 + 129}{16}, cos (x) = \frac{1 - 129}{16}

cos (x) = \frac{1 + 129}{16} : x = arccos (\frac{1 + 129}{16}) + 2 πn, x = 2 π - arccos (\frac{1 + 129}{16}) + 2 πn

cos (x) = \frac{1 + 129}{16}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1 + 129}{16}

以下の一般解

cos (x) = \frac{1 + 129}{16}

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

x = arccos (\frac{1 + 129}{16}) + 2 πn, x = 2 π - arccos (\frac{1 + 129}{16}) + 2 πn

x = arccos (\frac{1 + 129}{16}) + 2 πn, x = 2 π - arccos (\frac{1 + 129}{16}) + 2 πn

cos (x) = \frac{1 - 129}{16} : x = arccos (\frac{1 - 129}{16}) + 2 πn, x = - arccos (\frac{1 - 129}{16}) + 2 πn

cos (x) = \frac{1 - 129}{16}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1 - 129}{16}

以下の一般解

cos (x) = \frac{1 - 129}{16}

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

x = arccos (\frac{1 - 129}{16}) + 2 πn, x = - arccos (\frac{1 - 129}{16}) + 2 πn

x = arccos (\frac{1 - 129}{16}) + 2 πn, x = - arccos (\frac{1 - 129}{16}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{1 + 129}{16}) + 2 πn, x = 2 π - arccos (\frac{1 + 129}{16}) + 2 πn, x = arccos (\frac{1 - 129}{16}) + 2 πn, x = - arccos (\frac{1 - 129}{16}) + 2 πn

10進法形式で解を証明する

x = 0.68824 \dots + 2 πn, x = 2 π - 0.68824 \dots + 2 πn, x = 2.27491 \dots + 2 πn, x = - 2.27491 \dots + 2 πn

グラフ

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