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受欢迎的三角函数 >

cos^2(θ)-cos(θ)-12=0

解答

cos^{2} (θ) - cos (θ) - 12 = 0

解答

θ \in R 无解

求解步骤

cos^{2} (θ) - cos (θ) - 12 = 0

用替代法求解

cos^{2} (θ) - cos (θ) - 12 = 0

令

: cos (θ) = u

u^{2} - u - 12 = 0

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

u^{2} - u - 12 = 0

使用求根公式求解

u^{2} - u - 12 = 0

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 12 = 48

4 \cdot 1 \cdot 12

数字相乘:

4 \cdot 1 \cdot 12 = 48

= 48

= 1 + 48

数字相加:

1 + 48 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

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

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

使用法则

- (- a) = a

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

数字相减:

1 - 7 = - 6

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 6}{2}

使用分式法则:

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

= - \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 (θ)

代回

cos (θ) = 4, cos (θ) = - 3

cos (θ) = 4, cos (θ) = - 3

cos (θ) = 4 :

无解

cos (θ) = 4

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - 3 :

无解

cos (θ) = - 3

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ \in R 无解

作图

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