2sin2θ−3cos2θ−3sinθ+3=sinθ(4cosθ+6s inθ−3)

    left-hand side    use identities:     (using that form of cosine double angle because there's one 1 cosine on the right-hand side. if we used another form, there'd be too many cosines)    those +3 and -3 cancel each other out (add to become 0)    factor out  out of everything       

Ithink the answer is b