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 >

solvefor x,cos(2x)=(m-1)/(m+1)

解答

求解

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

解答

x = \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn, x = - \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn

求解步骤

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

使用反三角函数性质

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

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

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

2 x = arccos (\frac{m - 1}{m + 1}) + 2 πn, 2 x = - arccos (\frac{m - 1}{m + 1}) + 2 πn

2 x = arccos (\frac{m - 1}{m + 1}) + 2 πn, 2 x = - arccos (\frac{m - 1}{m + 1}) + 2 πn

解

2 x = arccos (\frac{m - 1}{m + 1}) + 2 πn : x = \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn

2 x = arccos (\frac{m - 1}{m + 1}) + 2 πn

两边除以

2

2 x = arccos (\frac{m - 1}{m + 1}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn

x = \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn

解

2 x = - arccos (\frac{m - 1}{m + 1}) + 2 πn : x = - \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn

2 x = - arccos (\frac{m - 1}{m + 1}) + 2 πn

两边除以

2

2 x = - arccos (\frac{m - 1}{m + 1}) + 2 πn

两边除以

2

\frac{2 x}{2} = - \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + \frac{2 πn}{2}

化简

x = - \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn

x = - \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn

x = \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn, x = - \frac{arccos ( \frac{m - 1}{m + 1} )}{2} + πn

作图

流行的例子

6 tan^{2} (x) + 8 = 10

2sin(x-pi/4)+sqrt(2)cos(x)=1

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

sin^2(θ)=2cos(θ)+1

sin^{2} (θ) = 2 cos (θ) + 1

sin(6x)cos(10x)-cos(6x)sin(10x)=-0.1

sin (6 x) cos (10 x) - cos (6 x) sin (10 x) = - 0.1

(cos(θ))(cos(θ)+1)=0

(cos (θ)) (cos (θ) + 1) = 0