Question

Solve equation for principle value of x Tan2x=cotx?

Answer 1

`x_1=-pi/2+2kpi,x_2=pi/2+2kpi,x_3-5/6*pi+2kpi,x_4=pi/6+2kpi,x_5=-pi/6+2kpi,x_6=5/6*pi+2kpi`

Explanation:

Using the Addition formula for the tan-function we get

`(2sin(x)/cos(x))/(1-sin^2(x)/cos^2(x))=cos(x)/sin(x)`
so we get

`(2sin(x)cos(x))/(cos^2(x)-sin^2(x))=cos(x)/sin(x)`
and this is

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