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受欢迎的三角函数 >

sqrt(3)cos(x)+sin(x)=2

解答

3 cos (x) + sin (x) = 2

解答

x = \frac{π}{6} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (x) + sin (x) = 2

两边减去

sin (x)

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

数字相加/相减:

- 4 + 3 = - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

用替代法求解

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

令

: sin (x) = u

- 1 + 4 u - 4 u^{2} = 0

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

- 1 + 4 u - 4 u^{2} = 0

改写成标准形式

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

- 4 u^{2} + 4 u - 1 = 0

使用求根公式求解

- 4 u^{2} + 4 u - 1 = 0

二次方程求根公式:

若

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

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

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

4^{2} - 4 (- 4) (- 1) = 0

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

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

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

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

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

去除括号:

(- a) = - a

= \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}

u = \frac{1}{2}

二次方程组的解是:

u = \frac{1}{2}

u = sin (x)

代回

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

将解代入原方程进行验证

将它们代入

3 cos (x) + sin (x) = 2

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{6} + 2 πn

的解:

真

\frac{π}{6} + 2 πn

代入

n = 1

\frac{π}{6} + 2 π 1

对于

3 cos (x) + sin (x) = 2

代入

x = \frac{π}{6} + 2 π 1

3 cos (\frac{π}{6} + 2 π 1) + sin (\frac{π}{6} + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

\frac{5 π}{6} + 2 πn

的解:

假

\frac{5 π}{6} + 2 πn

代入

n = 1

\frac{5 π}{6} + 2 π 1

对于

3 cos (x) + sin (x) = 2

代入

x = \frac{5 π}{6} + 2 π 1

3 cos (\frac{5 π}{6} + 2 π 1) + sin (\frac{5 π}{6} + 2 π 1) = 2

整理后得

- 1 = 2

\Rightarrow 假

x = \frac{π}{6} + 2 πn

作图

流行的例子

cos(2t)-cos(t)=-0.5

cos (2 t) - cos (t) = - 0.5

sin^2(x)-cos(x)+1=0

sin^{2} (x) - cos (x) + 1 = 0

tan(2x)=2sin(x)

tan (2 x) = 2 sin (x)

sin(θ)-cos(θ)=sqrt((2+\sqrt{3))/2}

sin (θ) - cos (θ) = \frac{2 + 3}{2}

2sin(2x)*cos(3x)+cos(3x)=0

2 sin (2 x) \cdot cos (3 x) + cos (3 x) = 0