Question

The number of solution(s) of equation ${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10,$ where $0\le x\le 2\pi$ is

Answer 1

${16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10$
$\Rightarrow {16}^{{\sin }^{2}x}+{16}^{1-{\sin }^{2}x}=10$
$\Rightarrow {16}^{{\sin }^{2}x}+\frac{16}{{16}^{{\sin }^{2}x}}=10$
Let ${16}^{{\sin }^{2}x}=t$
$\Rightarrow t+\frac{16}{t}=10$
$\Rightarrow t^2-10t+16=0$
$\Rightarrow (t-2)(t-8)=0$
$\Rightarrow t=2\text{or}t=8$

Now, ${16}^{{\sin }^{2}x}=2\text{or}{16}^{{\sin }^2}=8$
$\Rightarrow 2^{4{\sin }^{2}x}=2^1\text{or}{2}^{4{\sin }^{2}x}=2^3$
$\Rightarrow 4{\sin }^{2}x=1\text{or}4{\sin }^{2}x=3$
$\Rightarrow {\sin }^{2}x=\frac{1}{4}\text{or}{\sin }^{2}x=\frac{3}{4}$
$\Rightarrow \sin x=\pm \frac{1}{2}\text{or}\sin x=\pm \frac{\sqrt{3}}{2}$
$\Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}\text{or}x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$
Hence, number of solutions = 8