JEE Advance - Mathematics (1989 - No. 19)
The general solutions of $$\,\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$$ is
$$n\pi + {\pi \over 8}$$
$${{n\pi } \over 2} + {\pi \over 8}$$
$${\left( { - 1} \right)^n}{{n\pi } \over 2} + {\pi \over 8}\,\,$$
$$2n\pi + {\cos ^{ - 1}}{3 \over 2}$$