ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

4cos(2x)-3cos(x)+1=0

解

4 cos (2 x) - 3 cos (x) + 1 = 0

解

x = 0.59538 \dots + 2 πn, x = 2 π - 0.59538 \dots + 2 πn, x = 2.04085 \dots + 2 πn, x = - 2.04085 \dots + 2 πn

+1

度

x = 34.11286 \dots^{\circ} + 36 0^{\circ} n, x = 325.88713 \dots^{\circ} + 36 0^{\circ} n, x = 116.93210 \dots^{\circ} + 36 0^{\circ} n, x = - 116.93210 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

4 cos (2 x) - 3 cos (x) + 1 = 0

三角関数の公式を使用して書き換える

1 - 3 cos (x) + 4 cos (2 x)

2倍角の公式を使用:

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

= 1 - 3 cos (x) + 4 (2 cos^{2} (x) - 1)

簡素化

1 - 3 cos (x) + 4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) - 3 cos (x) - 3

1 - 3 cos (x) + 4 (2 cos^{2} (x) - 1)

拡張

4 (2 cos^{2} (x) - 1) : 8 cos^{2} (x) - 4

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

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (x), c = 1

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

簡素化

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

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

数を乗じる:

4 \cdot 2 = 8

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

数を乗じる:

4 \cdot 1 = 4

= 8 cos^{2} (x) - 4

= 8 cos^{2} (x) - 4

= 1 - 3 cos (x) + 8 cos^{2} (x) - 4

簡素化

1 - 3 cos (x) + 8 cos^{2} (x) - 4 : 8 cos^{2} (x) - 3 cos (x) - 3

1 - 3 cos (x) + 8 cos^{2} (x) - 4

条件のようなグループ

= - 3 cos (x) + 8 cos^{2} (x) + 1 - 4

数を足す/引く:

1 - 4 = - 3

= 8 cos^{2} (x) - 3 cos (x) - 3

= 8 cos^{2} (x) - 3 cos (x) - 3

= 8 cos^{2} (x) - 3 cos (x) - 3

- 3 - 3 cos (x) + 8 cos^{2} (x) = 0

置換で解く

- 3 - 3 cos (x) + 8 cos^{2} (x) = 0

仮定:

cos (x) = u

- 3 - 3 u + 8 u^{2} = 0

- 3 - 3 u + 8 u^{2} = 0 : u = \frac{3 + 105}{16}, u = \frac{3 - 105}{16}

- 3 - 3 u + 8 u^{2} = 0

標準的な形式で書く

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

8 u^{2} - 3 u - 3 = 0

解くとthe二次式

8 u^{2} - 3 u - 3 = 0

二次Equationの公式:

次の場合:

a = 8, b = - 3, c = - 3

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

(- 3)^{2} - 4 \cdot 8 (- 3) = 105

(- 3)^{2} - 4 \cdot 8 (- 3)

規則を適用

- (- a) = a

= (- 3)^{2} + 4 \cdot 8 \cdot 3

指数の規則を適用する:

n

が偶数であれば

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

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 8 \cdot 3

数を乗じる:

4 \cdot 8 \cdot 3 = 96

= 3^{2} + 96

3^{2} = 9

= 9 + 96

数を足す:

9 + 96 = 105

= 105

u_{1, 2} = \frac{- ( - 3 ) \pm 105}{2 \cdot 8}

解を分離する

u_{1} = \frac{- ( - 3 ) + 105}{2 \cdot 8}, u_{2} = \frac{- ( - 3 ) - 105}{2 \cdot 8}

u = \frac{- ( - 3 ) + 105}{2 \cdot 8} : \frac{3 + 105}{16}

\frac{- ( - 3 ) + 105}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{3 + 105}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{3 + 105}{16}

u = \frac{- ( - 3 ) - 105}{2 \cdot 8} : \frac{3 - 105}{16}

\frac{- ( - 3 ) - 105}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{3 - 105}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{3 - 105}{16}

二次equationの解:

u = \frac{3 + 105}{16}, u = \frac{3 - 105}{16}

代用を戻す

u = cos (x)

cos (x) = \frac{3 + 105}{16}, cos (x) = \frac{3 - 105}{16}

cos (x) = \frac{3 + 105}{16}, cos (x) = \frac{3 - 105}{16}

cos (x) = \frac{3 + 105}{16} : x = arccos (\frac{3 + 105}{16}) + 2 πn, x = 2 π - arccos (\frac{3 + 105}{16}) + 2 πn

cos (x) = \frac{3 + 105}{16}

三角関数の逆数プロパティを適用する

cos (x) = \frac{3 + 105}{16}

以下の一般解

cos (x) = \frac{3 + 105}{16}

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

x = arccos (\frac{3 + 105}{16}) + 2 πn, x = 2 π - arccos (\frac{3 + 105}{16}) + 2 πn

x = arccos (\frac{3 + 105}{16}) + 2 πn, x = 2 π - arccos (\frac{3 + 105}{16}) + 2 πn

cos (x) = \frac{3 - 105}{16} : x = arccos (\frac{3 - 105}{16}) + 2 πn, x = - arccos (\frac{3 - 105}{16}) + 2 πn

cos (x) = \frac{3 - 105}{16}

三角関数の逆数プロパティを適用する

cos (x) = \frac{3 - 105}{16}

以下の一般解

cos (x) = \frac{3 - 105}{16}

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

x = arccos (\frac{3 - 105}{16}) + 2 πn, x = - arccos (\frac{3 - 105}{16}) + 2 πn

x = arccos (\frac{3 - 105}{16}) + 2 πn, x = - arccos (\frac{3 - 105}{16}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{3 + 105}{16}) + 2 πn, x = 2 π - arccos (\frac{3 + 105}{16}) + 2 πn, x = arccos (\frac{3 - 105}{16}) + 2 πn, x = - arccos (\frac{3 - 105}{16}) + 2 πn

10進法形式で解を証明する

x = 0.59538 \dots + 2 πn, x = 2 π - 0.59538 \dots + 2 πn, x = 2.04085 \dots + 2 πn, x = - 2.04085 \dots + 2 πn

グラフ

人気の例

2tan^4(x)-tan^2(x)-15=0

2 tan^{4} (x) - tan^{2} (x) - 15 = 0

cos(x)=-sqrt(1/2)

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

6cos(θ)=-6(1+cos(θ))

6 cos (θ) = - 6 (1 + cos (θ))

2sin^2(x)+5sin(x)+3=0

2 sin^{2} (x) + 5 sin (x) + 3 = 0

cos (A) = \frac{4}{5}