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受欢迎的三角函数 >

cos^2(x)-7cos(x)+12=0

解答

cos^{2} (x) - 7 cos (x) + 12 = 0

解答

x \in R 无解

求解步骤

cos^{2} (x) - 7 cos (x) + 12 = 0

用替代法求解

cos^{2} (x) - 7 cos (x) + 12 = 0

令

: cos (x) = u

u^{2} - 7 u + 12 = 0

u^{2} - 7 u + 12 = 0 : u = 4, u = 3

u^{2} - 7 u + 12 = 0

使用求根公式求解

u^{2} - 7 u + 12 = 0

二次方程求根公式:

若

a = 1, b = - 7, c = 12

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 12 = 1

(- 7)^{2} - 4 \cdot 1 \cdot 12

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 1 \cdot 12

数字相乘:

4 \cdot 1 \cdot 12 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

数字相减:

49 - 48 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 1}, u_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 1}

u = \frac{- ( - 7 ) + 1}{2 \cdot 1} : 4

\frac{- ( - 7 ) + 1}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{7 + 1}{2 \cdot 1}

数字相加:

7 + 1 = 8

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

数字相乘:

2 \cdot 1 = 2

= \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= 4

u = \frac{- ( - 7 ) - 1}{2 \cdot 1} : 3

\frac{- ( - 7 ) - 1}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{7 - 1}{2 \cdot 1}

数字相减:

7 - 1 = 6

= \frac{6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

二次方程组的解是:

u = 4, u = 3

u = cos (x)

代回

cos (x) = 4, cos (x) = 3

cos (x) = 4, cos (x) = 3

cos (x) = 4 :

无解

cos (x) = 4

- 1 \leq cos (x) \leq 1

无解

cos (x) = 3 :

无解

cos (x) = 3

- 1 \leq cos (x) \leq 1

无解

合并所有解

x \in R 无解

作图

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