Show that 2sin(2x)-3cos(2x)-3sin(x)+3=sin(x)(4cos(x)+6sin(x)-3)

When proving trigonometric identites, we must show that the left hand side of the equation = the right hand side. Here we will start with the left hand side (LHS) and show that it is equivalent to the right hand side (RHS).LHS=2sin(2x)-3cos(2x)-3sin(x)+3Using the double angle rules for sin(2x) and cos(2x);LHS=2(2sin(x)cos(x))-3(cos²(x)-sin²(x))-3sin(x)+3Notice that the RHS has sin(x) factorised out, meaning that every term in the LHS has a common factor of sin(x). Currently the LHS has a cos²x term, but we can change this to a sin²x term using the identity: cos²(x)=1-sin²(x) LHS=2(2cos(x)sin(x))-3(1-sin²(x)-sin²(x))-3sin(x)+3=4cos(x)sin(x)-3(1-2sin²(x))-3sin(x)+3=4cos(x)sin(x)-3+6sin²(x)-3sin(x)+3=4cos(x)sin(x)+6sin²(x)-3sin(x)=sin(x)(4cos(x)+6sin(x)-3)=RHSWe have shown that LHS=RHS, therefore the proof is complete.