zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

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

使用三角恒等式改写

- 3 + sin^{2} (x) - 3 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)

= - 3 + 1 - cos^{2} (x) - 3 cos (x)

化简

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

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

用替代法求解

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

令

: cos (x) = u

- 2 - u^{2} - 3 u = 0

- 2 - u^{2} - 3 u = 0 : u = - 2, u = - 1

- 2 - u^{2} - 3 u = 0

改写成标准形式

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

- u^{2} - 3 u - 2 = 0

使用求根公式求解

- u^{2} - 3 u - 2 = 0

二次方程求根公式:

若

a = - 1, b = - 3, c = - 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

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

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

使用法则

- (- a) = a

= (- 3)^{2} - 4 \cdot 1 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

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

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

去除括号:

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

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

数字相加:

3 + 1 = 4

= \frac{4}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{- 2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

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

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

去除括号:

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

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

数字相减:

3 - 1 = 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 = - 2, u = - 1

u = cos (x)

代回

cos (x) = - 2, cos (x) = - 1

cos (x) = - 2, cos (x) = - 1

cos (x) = - 2 :

无解

cos (x) = - 2

- 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

csc^{2} (x) = 0

1 + 2 cos (θ) = 0

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

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

-3cos(2θ)+14sin(θ)+3=sin(θ)+8

- 3 cos (2 θ) + 14 sin (θ) + 3 = sin (θ) + 8

cos (\frac{π}{6} t) = 0