Question 677925
Solve for x in [0,2pi]:
cos 2x = -sin x
cos^2x-sin^2x=-sinx
1-sin^2x-sin^2x=-sinx
-2sin^2x+sinx+1=0
2sin^2x-sinx-1=0
(2sinx+1)(sinx-1)=0
..
2sinx+1=0
sinx=-1/2
x=7&#960;/6, 11&#960;/6 (in quadrants III and IV where sin<0)
and
sinx-1=0
sinx=1
x=&#960;/2