Trigonometry

[ritwik06]

1. The problem statement, all variables and given/known data

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


3. 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?????