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2(sinX-cos2X)-sin2X(1+2sinX)+2cosX=0 this is my question previous solution not for this
2(sinx-cos2x)-2sinxcosx-2sin^2xcosx+2cosx=02(sinx-cos2x)-sin2x(1+2sinx)+2cosx=0sinx-cos2x-sinxcosx-sin^2xcosx+cosx=0sinx-(1-2sin^2x)-sinxcosx-sin^2xcosx+cosx=0 {cos2x=1-2sin^2x sinx-1+2sin^2x-sinxcosx-sin^2xcosx+cosx=0 and sin2x=2sinxcosx}sinx-sinxcosx+2sin^2x-sin^2xcosx+cosx-1=0sinx(1-cosx)+2sin^2x(1-cosx)-(1-cosx)=0(1-cosx)[2sin^2x+sinx-1]=01-cosx=0 or 2sin^2x+sinx-1=0now 2sin^2x+sinx-1=0 sinx=[-1+- root{1-4*2(-1)}]/2*2 sinx=-1 and ½ now write general solution for cosx=1 ,sinx=-1 and sinx=1/2
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