∴ 3−3cosx=sinx Now squaring on both side, we get 9+9cos2x−18cosx=sin2x 9cos2x−18cosx+9=1−cos2x ∴10cos2x−18cosx+8=0 ∴ 5cos2x−9cosx+4=0 ∴5cos2x−5cosx−4cosx+4=0
∴5cosx(cosx−1)−4(cosx−1)=0
∴(5cosx−4)(cosx−1)=0 ∴ cosx=
4
5
orcosx=1 Now for cosx=
4
5
,sin2x=1−cos2x =1−
16
25
=
9
25
∴cos2x−sin2x=
16
25
−
9
25
=
7
25
and for cosx=1(Which is not possible as cosx−1 for x=0 ) Hence option (d) is correct.