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受欢迎的三角函数 >

cos^2(θ)+9cos(θ)+18=0

解答

cos^{2} (θ) + 9 cos (θ) + 18 = 0

解答

θ \in R 无解

求解步骤

cos^{2} (θ) + 9 cos (θ) + 18 = 0

用替代法求解

cos^{2} (θ) + 9 cos (θ) + 18 = 0

令

: cos (θ) = u

u^{2} + 9 u + 18 = 0

u^{2} + 9 u + 18 = 0 : u = - 3, u = - 6

u^{2} + 9 u + 18 = 0

使用求根公式求解

u^{2} + 9 u + 18 = 0

二次方程求根公式:

若

a = 1, b = 9, c = 18

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 18}{2 \cdot 1}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 18}{2 \cdot 1}

9^{2} - 4 \cdot 1 \cdot 18 = 3

9^{2} - 4 \cdot 1 \cdot 18

数字相乘:

4 \cdot 1 \cdot 18 = 72

= 9^{2} - 72

9^{2} = 81

= 81 - 72

数字相减:

81 - 72 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 9 \pm 3}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 9 + 3}{2 \cdot 1}, u_{2} = \frac{- 9 - 3}{2 \cdot 1}

u = \frac{- 9 + 3}{2 \cdot 1} : - 3

\frac{- 9 + 3}{2 \cdot 1}

数字相加/相减:

- 9 + 3 = - 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

u = \frac{- 9 - 3}{2 \cdot 1} : - 6

\frac{- 9 - 3}{2 \cdot 1}

数字相减:

- 9 - 3 = - 12

= \frac{- 12}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 12}{2}

使用分式法则:

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

= - \frac{12}{2}

数字相除:

\frac{12}{2} = 6

= - 6

二次方程组的解是:

u = - 3, u = - 6

u = cos (θ)

代回

cos (θ) = - 3, cos (θ) = - 6

cos (θ) = - 3, cos (θ) = - 6

cos (θ) = - 3 :

无解

cos (θ) = - 3

- 1 \leq cos (x) \leq 1

无解

cos (θ) = - 6 :

无解

cos (θ) = - 6

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ \in R 无解

作图

流行的例子

sin (3 x) + 1 = 0

(cot(x)-1)(sqrt(3)cot(x)+1)=0

(cot (x) - 1) (3 cot (x) + 1) = 0

sin(x)cos(x)=sin(x)

sin (x) cos (x) = sin (x)

csc (θ) - 2 = 0

2sin^2(θ)=3(1-cos(θ))

2 sin^{2} (θ) = 3 (1 - cos (θ))