The number of roots of the equation $3\,\,{\sin ^2}x = 8\,\,\cos \,\,x$ in $\left( {\; - \frac{{\rm{\pi }}}{2},\,\,\frac{{\rm{\pi }}}{2}} \right)$ is

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Application +4 | -1

The number of roots of the equation

3\,\,{\sin ^2}x = 8\,\,\cos \,\,x

\left( {\; - \frac{\mathrm{\pi }}{2},\,\,\frac{\mathrm{\pi }}{2}} \right)

View answer and explanation

Correct option: A

\mathrm{1}

Option A (

\mathrm{1}

) is correct because by solving the equation

3\,{\sin ^2}x = 8\,\cos \,x

in the interval

\left( {\; - \frac{\mathrm{\pi }}{2},\,\frac{\mathrm{\pi }}{2}} \right)

, we find that there is only one root.

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