Using sinx-siny=2cos(x+y)/2 .sin(x-y)/2
let be x=2x and y =x
then in lhs;
sin3x+sin2x-sonx
put the value of x and 2x that we assume
firsly is sin2x-sinx is in the form of sinx-siny so that use formula
sin3x+2cos(2x+x)/2.sin(2x-2)/2
=sin3x+2cos3x/2.sinx/2

{{we know that sin2x=2sinx.cosx
divide by x by x/2
sin2x/2=2sinx/2.cosx/2
sinx=2sinx/2.cosx/2
now x replace by 3x
sin3x=2sin3x/2.cos3x/2}}


then; 2sin3x/2.cos3x/2+(2cos3x/2.sinx/2)
2cos3x/2 {sin3x/2+sinx/2}

using sinx+siny = 2sin(x+y)/2.cos(x-y)/2
put x=3x/2 and y =x/2

2cos3x/2{2sin{3x/2+x/2).cos (3x/2-x/2)
on solving that equation we get

2cos3x/2{2sinx.cosx/2}
then we get finally
4cos 3x/2.sinx.cosx/2
we can also write
4sinx.cosx/2.cos3x/2

Lhs= rhs hence proved..