For $x \in(0, \pi)$, the equation $\sin x+2 \sin 2 x-\sin 3 x=3$ has

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Trigonometric Equations

For

x \in (0, π)

, the equation

sin x + 2 sin 2 x - sin 3 x = 3

has

Answer: (D)

Rewrite the equation:

sin x + 2 sin 2 x - sin 3 x = 3

2 sin 2 x - (sin 3 x - sin x) = 3

2 sin 2 x - 2 cos 2 x sin x = 3

2 sin x (2 cos x - cos 2 x) = 3

2 sin x (2 cos x - 2 cos^{2} x + 1) = 3

2 sin x (- 2 cos^{2} x + 2 cos x + 1) = 3

For

x \in (0, π)

sin x \in (0, 1]

with maximum 1.

The expression

- 2 cos^{2} x + 2 cos x + 1

achieves maximum

3/2

when

cos x = 1/2

.

Thus the LHS has maximum

2 \times 1 \times 3/2 = 3

, but this requires both

sin x = 1

and

cos x = 1/2

simultaneously, which is impossible.

Therefore, the equation has no solution.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option