Question

Find the total number of solutions of $16^{{\sin }^{2}x}+16^{{\cos }^{2}x}=10$ in $x\in \left[0,3\pi \right]$ .

Answer 1

Answer: 12

We have, $16^{{\sin }^{2}x}+16^{{\cos }^{2}x}=10$
$\Rightarrow 16^{{\sin }^{2}x}+16^{1-{\sin }^{2}x}=10$
$\Rightarrow 16^{si{n}^{2}x}+\frac{16}{16^{{\sin }^{2}x}}=10$
Let $16^{{\sin }^{2}x}=y$
$\therefore y+\frac{16}{y}=10$
$\Rightarrow y^2-10y+16=0$
$\Rightarrow \left(y-8\right)\left(y-2\right)=0$
$\Rightarrow y=8$ and $y=2$
$\therefore 16^{{\sin }^{2}x}=8\&16^{{\sin }^{2}x}=2$
Taking $lo{g}_2$ both the sides in the above equation we get,
${\sin }^{2}x=\frac{3}{4},{\sin }^{2}x=\frac{1}{4}$ .
$\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2},\pm \frac{1}{2}$
$\Rightarrow \sin x=\sin \left(\pm \frac{\pi }{3}\right)\&\sin x=\sin \left(\pm \frac{\pi }{6}\right)$
$\Rightarrow x=n\pi +{\left(-1\right)}^n\left(\pm \frac{\pi }{3}\right)\&n\pi +{\left(-1\right)}^n\left(\pm \frac{\pi }{6}\right),n\in I$
$\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{8\pi }{3},\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6},\frac{13\pi }{6},\frac{17}{6}\pi$
So, there are $12$ solutions in $[0,3\pi ].$