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JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

The number of solutions in [0,2 π] of the equation 16^ ^2 x +16^ ^2...

The number of solutions in

[0,2 \pi]

of the equation

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

is

Options

A2
B4
C6
D8

Correct answer

D. 8

Step-by-step solution

\begin{aligned} & 16^{\sin ^2 x}+16^{\cos ^2 x}=10 \\ & 16^{\sin ^2 x}+16^{1-\sin ^2 x}=10 \\ & 16^{\sin ^2 x}+\frac{16}{16^{\sin ^2 x}}=10\end{aligned}

Let

16^{\sin ^2 x}=\mathrm{t}

\begin{array}{ll} \therefore & \mathrm{t}+\frac{16}{\mathrm{t}}=10 \\ \therefore & \mathrm{t}^2-10 \mathrm{t}+16=0 \\ & \Rightarrow \mathrm{t}=2 \text { and } \mathrm{t}=8 \end{array}

Now,

16^{\sin ^2 x}=2

and

16^{\sin ^2

x}=8

\begin{aligned} & 2^{4 \sin ^2 x}=2^1 \text { and }2^{4 \sin ^2 x}=2^3 \\ \therefore \quad \text{ and } 4 \sin ^2 x=1 \text { and }4 \sin ^2 x=3 \\ \therefore \quad \text{ and } \sin ^2 x=\frac{1}{4} \text { and } \sin ^2 x=\frac{3}{4}\\ & \sin x= \pm \frac{1}{2} \text { and } \sinx= \pm \frac{\sqrt{3}}{2}\\ \therefore \quad & x=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6} \text { and }x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \end{aligned}

\therefore \quad number of solutions

=8

.

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