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cosxsinx=sin^2x, solve for x.
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\[cosxsinx= \sin^2x\], solve for x!
if you had xy = y^2 you would write it as y^2 -x y= 0- y( y-x)=0 and y=0 or y=x are the two answers can you use this idea to solve your problem?
cosxsinx-sin^2x=0 sinx(cosx-sinx)=0 sinx=0 cosx-sinx=0 x=pi, 0 x= ?
for the second equatin divide by cos x 1- sin x/ cos x =0 1 - tan x =0 tan x =1 so all angles that make tan x =1
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