JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Application +4 | -1

0 \leq x \leq \frac{\pi}{2}

and

81^{\sin ^2 x}+81^{\cos ^2 x}=30

, then

x

is equal to

View answer and explanation

Correct option: A

\frac{\pi}{6}

Option A (

\frac{{\pi}}{{6}}

) is correct because solving the equation

81^{\sin ^2 x}+81^{\cos ^2 x}=30

leads to

\sin x = \frac{{1}}{{2}}

\sin x = \frac{{\sqrt{{3}}}}{{2}}

, and considering the given range,

x = \frac{{\pi}}{{6}}

is a valid solution.

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