Question

The number of solutions of ${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10$, $0\le x\le 2\pi$

Answer 1

We have, ${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10$ (i)

If ${16}^{{\sin }^{2}x}=t$, then $t+\frac{16}{t}=10$

Then Eq. (i) become
$t^2-10t+16=0$

$t=2,8$

$\Rightarrow$ ${16}^{{\sin }^{2}x}={16}^{1/4}$ or ${16}^{3/4}$

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

Now $\sin x=\frac{1}{2}$, then $x=\frac{\pi }{6},\frac{5\pi }{6}$

$\sin x=-\frac{1}{2}$, then $x=\frac{7\pi }{6}$ or $\frac{11\pi }{6}$

$\sin x=\frac{\sqrt{3}}{2}$, then $x=\frac{\pi }{3},\frac{2\pi }{3}$

$\sin x=-\frac{\sqrt{3}}{2}$, then $x=\frac{4\pi }{3},\frac{5\pi }{3}$

Hence, there will be eight solutions in all.
Ans: 8