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受欢迎的三角函数 >

-sqrt(3)cos(x)-sin(x)=-sqrt(3)

解答

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

解答

x = 2 πn, x = \frac{π}{3} + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

sin (x)

- 3 cos (x) = - 3 + sin (x)

两边进行平方

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

两边减去

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

3 cos^{2} (x) - 3 + 23 sin (x) - sin^{2} (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

- 3 - sin^{2} (x) + 3 (1 - sin^{2} (x)) + 2 sin (x) 3 : 23 sin (x) - 4 sin^{2} (x)

- 3 - sin^{2} (x) + 3 (1 - sin^{2} (x)) + 2 sin (x) 3

= - 3 - sin^{2} (x) + 3 (1 - sin^{2} (x)) + 23 sin (x)

乘开

3 (1 - sin^{2} (x)) : 3 - 3 sin^{2} (x)

3 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

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

数字相乘:

3 \cdot 1 = 3

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

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

化简

- 3 - sin^{2} (x) + 3 - 3 sin^{2} (x) + 2 sin (x) 3 : 23 sin (x) - 4 sin^{2} (x)

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

对同类项分组

= - sin^{2} (x) - 3 sin^{2} (x) + 23 sin (x) - 3 + 3

同类项相加:

- sin^{2} (x) - 3 sin^{2} (x) = - 4 sin^{2} (x)

= - 4 sin^{2} (x) + 23 sin (x) - 3 + 3

- 3 + 3 = 0

= 23 sin (x) - 4 sin^{2} (x)

= 23 sin (x) - 4 sin^{2} (x)

= 23 sin (x) - 4 sin^{2} (x)

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

用替代法求解

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

令

: sin (x) = u

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

- 4 u^{2} + 2 u 3 = 0 : u = 0, u = \frac{3}{2}

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

改写成标准形式

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

- 4 u^{2} + 23 u = 0

使用求根公式求解

- 4 u^{2} + 23 u = 0

二次方程求根公式:

若

a = - 4, b = 23, c = 0

u_{1, 2} = \frac{- 2 3 \pm ( 2 3 ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

u_{1, 2} = \frac{- 2 3 \pm ( 2 3 ) ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

(23)^{2} - 4 (- 4) \cdot 0 = 23

(23)^{2} - 4 (- 4) \cdot 0

使用法则

- (- a) = a

= (23)^{2} + 4 \cdot 4 \cdot 0

(23)^{2} = 2^{2} \cdot 3

(23)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

(3)^{2} : 3

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

约分:

2

= 1

= 3

= 2^{2} \cdot 3

4 \cdot 4 \cdot 0 = 0

4 \cdot 4 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 2^{2} \cdot 3 + 0

2^{2} \cdot 3 + 0 = 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= 3 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

2^{2} = 2

= 23

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

将解分隔开

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

u = \frac{- 2 3 + 2 3}{2 ( - 4 )} : 0

\frac{- 2 3 + 2 3}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 3 + 2 3}{- 2 \cdot 4}

同类项相加:

- 23 + 23 = 0

= \frac{0}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{0}{- 8}

使用分式法则:

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

= - \frac{0}{8}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

\frac{- 2 3 - 2 3}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 2 3 - 2 3}{- 2 \cdot 4}

同类项相加:

- 23 - 23 = - 43

= \frac{- 4 3}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 4 3}{- 8}

使用分式法则:

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

= \frac{4 3}{8}

约分:

4

= \frac{3}{2}

二次方程组的解是:

u = 0, u = \frac{3}{2}

u = sin (x)

代回

sin (x) = 0, sin (x) = \frac{3}{2}

sin (x) = 0, sin (x) = \frac{3}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = \frac{3}{2} : x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

sin (x) = \frac{3}{2}

sin (x) = \frac{3}{2}

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

- 3 cos (2 π 1) - sin (2 π 1) = - 3

整理后得

- 1.73205 \dots = - 1.73205 \dots

\Rightarrow 真

检验

π + 2 πn

的解:

假

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

- 3 cos (π + 2 π 1) - sin (π + 2 π 1) = - 3

整理后得

1.73205 \dots = - 1.73205 \dots

\Rightarrow 假

检验

\frac{π}{3} + 2 πn

的解:

真

\frac{π}{3} + 2 πn

代入

n = 1

\frac{π}{3} + 2 π 1

对于

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

代入

x = \frac{π}{3} + 2 π 1

- 3 cos (\frac{π}{3} + 2 π 1) - sin (\frac{π}{3} + 2 π 1) = - 3

整理后得

- 1.73205 \dots = - 1.73205 \dots

\Rightarrow 真

检验

\frac{2 π}{3} + 2 πn

的解:

假

\frac{2 π}{3} + 2 πn

代入

n = 1

\frac{2 π}{3} + 2 π 1

对于

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

代入

x = \frac{2 π}{3} + 2 π 1

- 3 cos (\frac{2 π}{3} + 2 π 1) - sin (\frac{2 π}{3} + 2 π 1) = - 3

整理后得

0 = - 1.73205 \dots

\Rightarrow 假

x = 2 πn, x = \frac{π}{3} + 2 πn

作图

流行的例子

9 cot^{2} (x) = 3

cos(x)=-sqrt(2)

cos (x) = - 2

tan(x)(cot(x)-1)=0

tan (x) (cot (x) - 1) = 0

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

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

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

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