Posted by Emma on .
2sinxcosx+4sin^2xcos^2x=0
solve for x in radians between [0,2pi)
(I mean that it is sine squared x and cosine squared x not sine to the power of 2x or cosine to the power of 2x)

trig 
Steve,
factor out sin*cos to get
2sin*cos(1+2sin*cos) = 0
so,
sin=0: x = 0 or π
or
cos=0: x = π/2 or 3π/2
or
1+sin2x = 0
sin2x = 1
2x = 3π/2: x = 3π/4