solve the equation for 0 < x < 2 Pi

2cos^2 (x) + sinx = 1

2cos^2 (x) = 0

sin (x) = 0

For the first one, try rewriting as:

2(1-sin^{2}(x))+sin(x)=1

2-2sin^{2}(x)+sin(x)-1=0

Now, let u=sin(x)

-2u^{2}+u+1=0

2u^{2}-u-1=0

Now solve by factoring or some other method and resub.