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受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

3 cos (x)

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

两边进行平方

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

两边减去

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

sin^{2} (x) - 3 - 6 cos (x) - 3 cos^{2} (x) = 0

使用三角恒等式改写

- 3 + sin^{2} (x) - 3 cos^{2} (x) - 6 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^{2} (x) - 6 cos (x)

化简

- 3 + 1 - cos^{2} (x) - 3 cos^{2} (x) - 6 cos (x) : - 4 cos^{2} (x) - 6 cos (x) - 2

- 3 + 1 - cos^{2} (x) - 3 cos^{2} (x) - 6 cos (x)

同类项相加:

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

= - 3 + 1 - 4 cos^{2} (x) - 6 cos (x)

数字相加/相减:

- 3 + 1 = - 2

= - 4 cos^{2} (x) - 6 cos (x) - 2

= - 4 cos^{2} (x) - 6 cos (x) - 2

- 2 - 4 cos^{2} (x) - 6 cos (x) = 0

用替代法求解

- 2 - 4 cos^{2} (x) - 6 cos (x) = 0

令

: cos (x) = u

- 2 - 4 u^{2} - 6 u = 0

- 2 - 4 u^{2} - 6 u = 0 : u = - 1, u = - \frac{1}{2}

- 2 - 4 u^{2} - 6 u = 0

改写成标准形式

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

- 4 u^{2} - 6 u - 2 = 0

使用求根公式求解

- 4 u^{2} - 6 u - 2 = 0

二次方程求根公式:

若

a = - 4, b = - 6, c = - 2

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 4 ) ( - 2 )}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 4 ) ( - 2 )}{2 ( - 4 )}

(- 6)^{2} - 4 (- 4) (- 2) = 2

(- 6)^{2} - 4 (- 4) (- 2)

使用法则

- (- a) = a

= (- 6)^{2} - 4 \cdot 4 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

数字相减:

36 - 32 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- ( - 6 ) + 2}{2 ( - 4 )}, u_{2} = \frac{- ( - 6 ) - 2}{2 ( - 4 )}

u = \frac{- ( - 6 ) + 2}{2 ( - 4 )} : - 1

\frac{- ( - 6 ) + 2}{2 ( - 4 )}

去除括号:

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

= \frac{6 + 2}{- 2 \cdot 4}

数字相加:

6 + 2 = 8

= \frac{8}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{8}{- 8}

使用分式法则:

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

= - \frac{8}{8}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 6 ) - 2}{2 ( - 4 )} : - \frac{1}{2}

\frac{- ( - 6 ) - 2}{2 ( - 4 )}

去除括号:

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

= \frac{6 - 2}{- 2 \cdot 4}

数字相减:

6 - 2 = 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{- 8}

使用分式法则:

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

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

二次方程组的解是:

u = - 1, u = - \frac{1}{2}

u = cos (x)

代回

cos (x) = - 1, cos (x) = - \frac{1}{2}

cos (x) = - 1, cos (x) = - \frac{1}{2}

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

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - \frac{1}{2}

cos (x) = - \frac{1}{2}

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

合并所有解

x = π + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

sin (π + 2 π 1) - 3 cos (π + 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:

真

\frac{2 π}{3} + 2 πn

代入

n = 1

\frac{2 π}{3} + 2 π 1

对于

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

代入

x = \frac{2 π}{3} + 2 π 1

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

整理后得

1.73205 \dots = 1.73205 \dots

\Rightarrow 真

检验

\frac{4 π}{3} + 2 πn

的解:

假

\frac{4 π}{3} + 2 πn

代入

n = 1

\frac{4 π}{3} + 2 π 1

对于

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

代入

x = \frac{4 π}{3} + 2 π 1

sin (\frac{4 π}{3} + 2 π 1) - 3 cos (\frac{4 π}{3} + 2 π 1) = 3

整理后得

0 = 1.73205 \dots

\Rightarrow 假

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

作图

流行的例子

cot^2(x)-2csc(x)=-1

cot^{2} (x) - 2 csc (x) = - 1

sqrt(3)csc(x)+csc(x)cot(x)=0

3 csc (x) + csc (x) cot (x) = 0

-3cos^2(θ)-10cos(θ)=-3cos(θ)-6

- 3 cos^{2} (θ) - 10 cos (θ) = - 3 cos (θ) - 6

-4sin(x)=-cos^2(x)+4

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

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

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