Substituting
cos 2x=cos^2 x-sin^2 x=2cos^2 x -1
And
sin 2x=2sin x*cos x
The expression becomes
rarr1-2cos^2 x+1+(sin x/cos x)/(1-sinx/cos x)=1+2sinx*cosx
rarr2-2cos^2 x+(sinx/cancel(cosx))(cancel(cosx)/(cosx-sin x))-1-
2sinx*cosx=0
rarr2sin^2 x(cosx-sinx)+sinx-(1+2sinx*cosx)(cosx-sinx)=0
rarr2sin^2 x*cosx-2sin^3 x+sin x -cosx+sinx-2sinx*cos^2 x+2sin^2
x*cosx=0
rarr4sin^2 x*cosx-2sin^3 x+2sinx-cosx-2sinx*cos^2x=0
rarr4sin^2 x*cosx-2sin^3 x+2sinx-cosx-2sinx*(1-sin^2 x)=0
rarr4sin^2 x*cosx-cancel(2sin^3 x)+cancel(2sinx)-cosx-
cancel(2sinx)+cancel(2sin^3 x)=0
rarr4sin^2 x*cosx-cosx=0
rarrcosx(4sin^2x-1)=0
cos x=0 => x=90^@ +k*180^@ -> but in the original expression there's tgx and tgx is undefined for x=90^@ or x=270^@ so we must outrightly reject this solution or only admit it when x->90^@
4sin^2 x-1=0 => sin^2 x=1/4 => sin x=1/2 => x=30^@+k*360^@ or x=150^@+k*360^@, where k in NN