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受欢迎的三角函数 >

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

解答

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

解答

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

度数

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

求解步骤

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

化简

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

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

数字相乘:

4 \cdot 2 = 8

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

数字相乘:

4 \cdot 1 = 4

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

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

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

化简

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

对同类项分组

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

数字相加/相减:

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

用替代法求解

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

令

: 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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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)

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 8 \cdot 3 = 96

= 3^{2} + 96

3^{2} = 9

= 9 + 96

数字相加:

9 + 96 = 105

= 105

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

将解分隔开

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}

使用法则

- (- a) = a

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

数字相乘:

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}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 8 = 16

= \frac{3 - 105}{16}

二次方程组的解是:

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

u = cos (x)

代回

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}

使用反三角函数性质

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

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

的通解

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}

使用反三角函数性质

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

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

的通解

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

合并所有解

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

以小数形式表示解

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

作图

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