The number of solutions of the equation `sin x . Sin 2x. Sin 3x=1` in `[0,2pi]` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

A

Explanation

`sin x.sin 2x.sin 3x=1`
Case I : sin x = 1 and sin 2x = -1 and sin 3x = -1
This does not hold simultaneously, because when sin x = -1, the value of cos x = -1/2
Case II : sin x=-1 and sin 2x = 1 and sin 3x = -1
This does not hold simultaneously, because when sin x = -1, the value of `sin 3x = 3 sin x - 4 sin^(3) x=1`
Case III : sin x = -1 and sin 2x = -1 and sin 3x = 1
This is not possible, because sin x and 2x cannot be simultaneously -1
Case IV : sin x, sin 2x, sin 3x = 1
Clearly, this does not hold simultaneously.