No. of solutions of ${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10,0\le x\le 2\pi$  is

${16}^{{\sin }^{2}x}+{16}^{1-{\sin }^{2}x}=10$

Let ${16}^{{\sin }^{2}x}=a$

$a+\frac{16}{a}=10$

$a^2+16-10a=0$

$a=2,8$

$a=2,a=8$

${16}^{{\sin }^{2}x}=2,{16}^{{\sin }^{2}x}=2^3$

${\sin }^{2}x=\frac{1}{4},{\sin }^{2}x=\frac{3}{4}$

$\sin x=\pm \frac{1}{2},\sin x=\pm \frac{\sqrt{3}}{2}$

$\therefore 8$  solutions. $\left(\because 0\le x\le 2\pi \sin x=\pm \frac{1}{2}has4solutions,\sin x=\pm \frac{\sqrt{3}}{2}has4solutions\right)$