JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Application +4 | -1

The number of solutions of

16^{\sin ^2 x}+16^{\cos ^2 x}=10,0 \leq x \leq 2 \pi

View answer and explanation

Correct option: A

Option A (8) is correct because solving the equation

16^{\sin^2 x} + 16^{\cos^2 x} = 10

leads to

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

and

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

, which have a total of 8 solutions in the interval

[0, 2\pi]

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