The number of solutions of the equation $1 + \sin x\,\,{\sin ^2}\frac{x}{2} = 0$ in $\left[ { - \,{\rm{\pi }},\,{\rm{\pi }}} \right]$ is

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

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Analyzing +4 | -1

The number of solutions of the equation

1 + \sin x\,\,{\sin ^2}\frac{x}{2} = 0

\left[ { - \,\mathrm{\pi },\,\mathrm{\pi }} \right]

View answer and explanation

Correct option: A

zero

Option A (zero) is correct because the equation

1 + \sin x \sin^2 \frac{x}{2} = 0

simplifies to

\sin 2x - 2 \sin x = 4

, which has no solutions since the range of

\sin

is [-1, 1].

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