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受欢迎的三角函数 >

sin^2(x)-8cos(x)-8=0

解答

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

解答

x = π + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

使用三角恒等式改写

- 8 + sin^{2} (x) - 8 cos (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

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

化简

= - cos^{2} (x) - 8 cos (x) - 7

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

用替代法求解

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

令

: cos (x) = u

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

- 7 - u^{2} - 8 u = 0 : u = - 7, u = - 1

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

改写成标准形式

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

- u^{2} - 8 u - 7 = 0

使用求根公式求解

- u^{2} - 8 u - 7 = 0

二次方程求根公式:

若

a = - 1, b = - 8, c = - 7

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 1 ) ( - 7 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 1 ) ( - 7 )}{2 ( - 1 )}

(- 8)^{2} - 4 (- 1) (- 7) = 6

(- 8)^{2} - 4 (- 1) (- 7)

使用法则

- (- a) = a

= (- 8)^{2} - 4 \cdot 1 \cdot 7

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 1 \cdot 7

数字相乘:

4 \cdot 1 \cdot 7 = 28

= 8^{2} - 28

8^{2} = 64

= 64 - 28

数字相减:

64 - 28 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- ( - 8 ) \pm 6}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 6}{2 ( - 1 )}, u_{2} = \frac{- ( - 8 ) - 6}{2 ( - 1 )}

u = \frac{- ( - 8 ) + 6}{2 ( - 1 )} : - 7

\frac{- ( - 8 ) + 6}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 + 6}{- 2 \cdot 1}

数字相加:

8 + 6 = 14

= \frac{14}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{14}{- 2}

使用分式法则:

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

= - \frac{14}{2}

数字相除:

\frac{14}{2} = 7

= - 7

u = \frac{- ( - 8 ) - 6}{2 ( - 1 )} : - 1

\frac{- ( - 8 ) - 6}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 - 6}{- 2 \cdot 1}

数字相减:

8 - 6 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{- 2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = - 7, u = - 1

u = cos (x)

代回

cos (x) = - 7, cos (x) = - 1

cos (x) = - 7, cos (x) = - 1

cos (x) = - 7 :

无解

cos (x) = - 7

- 1 \leq cos (x) \leq 1

无解

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

合并所有解

x = π + 2 πn

作图

流行的例子

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tan (\frac{1}{2} x) + 3 = 0

-arcsin(4x)= pi/4

- arcsin (4 x) = \frac{π}{4}