Solve sin2xcosx =cosx

I have a test tonight, and I don't know how to do this.
Please help!

sin(2x) cos(x) = cos(x)

=> sin(2x) cos(x) - cos(x) = 0

=> cos(x) [ sin(2x) - 1 ] = 0

case 1:

cos(x) = 0

x = π/2, 3π/2

case 2: sin(2x) - 1 = 0

=> sin(2x) = 1

=> 2x = π/2, 5π/2

=> x = π/4, 5π/4

x = π/4, π/2, 5π/4 and 3π/2 in the interval [0, 2π ]

sin^2xcosx = cosx
sin^2xcosx - cosx = 0
cosx(sin^2x - 1) = 0
cosx = 0 or sin^2x - 1 = 0
cosx = 0 or sin^2x = 1
cosx = 0 or sinx = 1 or sinx = -1
x = pi/2 + npi or x = pi/2 + 2npi or x = 3pi/2 + 2npi for all integers, n