Solve the equation `2 (cos x+cos 2x)+sin 2x (1+2 cos x)=2 sin x` for `x in [-pi, pi]`.

Explanation

The given equation is
`2(cos x+2 cos^(2) x-1)+2 sin x cos x (1+2 cos x) -2 sin x=0`
`rArr 2(2 cos^(2) x+ cos x-1)+2 sin x(2 cos^(2) x+ cos x -1)=0`
`rArr (sin x+1) (2 cos^(2) x+ cos x-1) =0`
`rArr sin x=-1 or 2 cos^(2) x+2 cos x- cos x-1 =0`
`rArr x=2npi -pi/2, n in Z or (2 cos x-1) (cos x+1)=0`
`rArr x=2n pi - pi/2, n in Z or x=2npi pm pi/3 or x=2n pi + pi, n in Z`
For `-pi le x le pi, x= -pi/2, pm pi/3, pm pi`