Consider the given equation. sinx−sin2x+sin3x=2cos2x−2cosx It can be solved as, 2sin2xcosx−sin2x=2cosx(cosx−1) 2sinxcosx(2cosx−1)=2cosx(cosx−1) So either, cosx=0 x=cos−1(0) =
π
2
……. (I) Or, sinx(2cosx−1)=cosx−1 2sinxcosx+1=sinx+cosx 4sin2xcos2x+1+4sinxcosx=1+2sinxcosx 2sinxcosx(sin2x+1)=0 Therefore, either sinx=0,cosx=0 or sin2x=−1 Now, x∈(0,π) So, sinx≠0 Now, sin2x=−1 2x=