Trigonometric Equation Simplification in the Interval [-pi, pi]

[ritwik06]

Homework Statement

Find the solution of
2(cos x + cos 2x)+sin 2x(1+2cos x)=2 sin x in the interval [-pi, pi]

The Attempt at a Solution

Simplifying:
2 (cos x- sin x)+ sin 2x +cos 2x +2sin 2x cos x=0
2 (cos x- sin x)+ sin 2x +cos 2x +sin 3x+sin x=0
cos 2x+2cos x+sin 2x-sin x+sin 3x=0

But now its difficult to convert into factors to apply zero prouct rule.

 

[HallsofIvy]

I would recommend you get rid of "sin 2x" and "cos 2x" by using sin 2x= 2sin x cos x and cos 2x= cos² x- sin² x.

 

[ritwik06]

I would recommend you get rid of "sin 2x" and "cos 2x" by using sin 2x= 2sin x cos x and cos 2x= cos² x- sin² x.

Changing all the multiples of x, I get:
cos²x+2 cos x +2 sin x cos x -4 sin³x+2sin x -sin ²x=0

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

Again I am stuck?