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受欢迎的三角函数 >

sin^2(x)-5cos(x)-5=0

解答

sin^{2} (x) - 5 cos (x) - 5 = 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) - 5 cos (x) - 5 = 0

使用三角恒等式改写

- 5 + sin^{2} (x) - 5 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)

= - 5 + 1 - cos^{2} (x) - 5 cos (x)

化简

= - cos^{2} (x) - 5 cos (x) - 4

- 4 - cos^{2} (x) - 5 cos (x) = 0

用替代法求解

- 4 - cos^{2} (x) - 5 cos (x) = 0

令

: cos (x) = u

- 4 - u^{2} - 5 u = 0

- 4 - u^{2} - 5 u = 0 : u = - 4, u = - 1

- 4 - u^{2} - 5 u = 0

改写成标准形式

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

- u^{2} - 5 u - 4 = 0

使用求根公式求解

- u^{2} - 5 u - 4 = 0

二次方程求根公式:

若

a = - 1, b = - 5, c = - 4

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 1 ) ( - 4 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 1 ) ( - 4 )}{2 ( - 1 )}

(- 5)^{2} - 4 (- 1) (- 4) = 3

(- 5)^{2} - 4 (- 1) (- 4)

使用法则

- (- a) = a

= (- 5)^{2} - 4 \cdot 1 \cdot 4

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 1 \cdot 4

数字相乘:

4 \cdot 1 \cdot 4 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 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{- ( - 5 ) \pm 3}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 3}{2 ( - 1 )}, u_{2} = \frac{- ( - 5 ) - 3}{2 ( - 1 )}

u = \frac{- ( - 5 ) + 3}{2 ( - 1 )} : - 4

\frac{- ( - 5 ) + 3}{2 ( - 1 )}

去除括号:

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

= \frac{5 + 3}{- 2 \cdot 1}

数字相加:

5 + 3 = 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

u = \frac{- ( - 5 ) - 3}{2 ( - 1 )} : - 1

\frac{- ( - 5 ) - 3}{2 ( - 1 )}

去除括号:

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

= \frac{5 - 3}{- 2 \cdot 1}

数字相减:

5 - 3 = 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 = - 4, u = - 1

u = cos (x)

代回

cos (x) = - 4, cos (x) = - 1

cos (x) = - 4, cos (x) = - 1

cos (x) = - 4 :

无解

cos (x) = - 4

- 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

(1+tan^2(27*θ))/(1-tan^2(27*θ))=sqrt(2)

\frac{1 + tan ^{2} ( 27 \cdot θ )}{1 - tan ^{2} ( 27 \cdot θ )} = 2

tan (x) = - cot (x)

3cos^2(x)-sin^2(x)=0

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

cos (x) = - 0.6

cos (x) = - 0.9