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

\(\begin{aligned} & 16^{\sin ^2 x}+16^{\cos ^2 x}=10 \\ & 16^{\sin ^2 x}+16^{1-\sin ^2 x}=10 \\ & 16^{\sin ^2 x}+\frac{16}{16^{\sin ^2 x}}=10\end{aligned}\)
Let \(16^{\sin ^2 x}=\mathrm{t}\)
\(\therefore \mathrm{t}+\frac{16}{\mathrm{t}}=10 \)
\( \therefore \mathrm{t}^2-10 \mathrm{t}+16=0 \)
\( \Rightarrow \mathrm{t}=2 \text { and } \mathrm{t}=8\)
Now, \(16^{\sin ^2 x}=2\) and \(16^{\sin ^2 x}=8\)
\(2^{4 \sin ^2 x}=2^1 \text { and } 2^{4 \sin ^2 x}=2^3 \)
\( \therefore 4 \sin ^2 x=1 \text { and } 4 \sin ^2 x=3 \)
\( \therefore \sin ^2 x=\frac{1}{4} \text { and } \sin ^2 x=\frac{3}{4} \)
\( \sin x= \pm \frac{1}{2} \text { and } \sin x= \pm \frac{\sqrt{3}}{2} \)
\( \therefore x=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6} \text { and } x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3}\)
\(\therefore\) number of solutions \(=8\).