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受欢迎的三角函数 >

cos^2(θ)-cos(θ)-20=0

解答

cos^{2} (θ) - cos (θ) - 20 = 0

解答

θ \in R 无解

求解步骤

cos^{2} (θ) - cos (θ) - 20 = 0

用替代法求解

cos^{2} (θ) - cos (θ) - 20 = 0

令

: cos (θ) = u

u^{2} - u - 20 = 0

u^{2} - u - 20 = 0 : u = 5, u = - 4

u^{2} - u - 20 = 0

使用求根公式求解

u^{2} - u - 20 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 20

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 20) = 9

(- 1)^{2} - 4 \cdot 1 \cdot (- 20)

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 20

(- 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 20 = 80

4 \cdot 1 \cdot 20

数字相乘:

4 \cdot 1 \cdot 20 = 80

= 80

= 1 + 80

数字相加:

1 + 80 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 1 ) \pm 9}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 9}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 9}{2 \cdot 1}

u = \frac{- ( - 1 ) + 9}{2 \cdot 1} : 5

\frac{- ( - 1 ) + 9}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 9}{2 \cdot 1}

数字相加:

1 + 9 = 10

= \frac{10}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{10}{2}

数字相除:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 1 ) - 9}{2 \cdot 1} : - 4

\frac{- ( - 1 ) - 9}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 9}{2 \cdot 1}

数字相减:

1 - 9 = - 8

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 8}{2}

使用分式法则:

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

= - \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= - 4

二次方程组的解是:

u = 5, u = - 4

u = cos (θ)

代回

cos (θ) = 5, cos (θ) = - 4

cos (θ) = 5, cos (θ) = - 4

cos (θ) = 5 :

无解

cos (θ) = 5

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - 4 :

无解

cos (θ) = - 4

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ \in R 无解

作图

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