The number of solutions in $[0,2 \pi]$ of the equation $16^{\sin ^2 x}+16^{\cos ^2 x}=10$ is

The number of solutions in $[0,2 \pi]$ of the equation $16^{\sin ^2 x}+16^{\cos ^2 x}=10$ is
  1. 2
  2. 4
  3. 6
  4. 8

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 $16^{\sin ^2 x}=8$ $\begin{aligned} & 2^{4 \sin ^2 x}=2^1 \text { and } 2^{4 \sin ^2 x}=2^3 \\ \therefore \quad & 4 \sin ^2 x=1 \text { and } 4 \sin ^2 x=3 \\ \therefore \quad & \sin ^2 x=\frac{1}{4} \text { and } \sin ^2 x=\frac{3}{4} \\ & \sin x= \pm \frac{1}{2} \text { and } \sin x= \pm \frac{\sqrt{3}}{2} \\ \therefore \quad & x=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6} \text { and } x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \end{aligned}$ $\therefore \quad$ number of solutions $=8$.

Asked in: MHT CET 2023 (09 May Shift 1)

Practice more Functions questions on Aicharya