Question 971540
Solve on the interval [0, 2pi]
Sin*2x - sin x + 1= cos*2x
sin^2(x)-sinx+1=1-sin^2(x)
2sin^2(x)-sinx=0
sinx(2sinx-1)=0
sinx=0
x=0, π, 2π
or 
2sinx-1=0
sinx=1/2
x=π/6, 11π/6