sin4x - sin2x = 0 solve for interval (0 2π)

Question

sin4x - sin2x = 0 solve for interval (0 2π)

algebraic means without calculator 

Answer 1

sin4x=2sin2xcos2x

cos2x=2cos²x-1

sin4x - sin2x = 2sin2xcos2x - sin2x=sin2x[2(2cos²x-1)-1] = sin2x(4cos²x-3)=0

 

sin2x=0 2sinxcosx=0 sinx  = 0 or cosx = 0, x = 0, pi/2, pi, 3pi/2, 2pi

 

or

 

4cos²x-3=0 cosx = ±root3/2, so x = pi/6, 5pi/6, 7pi/6, 11pi/6