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

We have, $16^{\sin^{2} x}+16^{\cos^{2} x}=10$
$\Rightarrow 16^{\sin^{2} x}+16^{1 - \sin^{2} x}=10$
$\Rightarrow 16^{sin^{2} x}+\frac{16}{16^{\sin^{2} x}}=10$
Let $16^{\sin^{2} x}=y$
$\therefore y+\frac{16}{y}=10$
$\Rightarrow y^{2}-10y+16=0$
$\Rightarrow \left(y - 8\right)\left(y - 2\right)=0$
$\Rightarrow y=8$ and $y=2$
$\therefore 16^{\sin^{2} x}=8\&16^{\sin^{2} x}=2$
Taking $log_{2}$ 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 $\left[\right.0,3\pi \left]\right..$