Question 733332
Solve 2 sin^2 x + cos x -2=0 on the interval 0 < x < 2&#960;
2 sin^2 x + cos x -2=0
2(1-cos^2x)+cosx-2=0
2-2cos^2x+cosx-2=0
-2cos^2x+cosx=0
cosx(1-2cosx)=0
cosx=0
x=&#960;/2, 3&#960;/2
or
1-2cosx=0
2cosx=1
cosx=1/2
x=&#960;/6, 11&#960;/6 (in Q1 and Q4 where cos>0)