Question 821209
2sin^2xcosx - cosx = 0 solve on the interval [0,2pi]
cosx(2sin^2x-1)=0
..
cosx=0
x=π/2, 3π/2
or
2sin^2x-1=0
sin^2x=1/2
sinx=±√(1/2)=±√2/2
x=π/4, 3π/4, 5π/4, 7π/4