Question

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

Answer 1

Let $16^{\sin ^{2} x}=y$
$16^{\cos ^{2} x}=16^{1-\sin ^{2} x}=16 y^{-1}$
The equation can be written as $y +\frac{16}{ y }=10$
$\Rightarrow y^{2}-10 y+16=0 $
$\Rightarrow (y-8)(y-2)=0 $
$ y=8, y=2$
If $ y=8$
$\Rightarrow 16^{\sin ^{2} x}=2^{3} $
$\Rightarrow 2^{4 \sin ^{2} x}=2^{3} $
$\Rightarrow \sin ^{2} x=\frac{3}{4} $
$\Rightarrow \sin ^{2} x=\pm \sin ^{2} \pi / 3$
$\Rightarrow x=n \pi \pm \pi / 3$
Which gives $3$ solutions in $[0,3 \pi]$
If $ y=2 $
$\Rightarrow 16^{\sin ^{2} x}=2 $
$ \Rightarrow 4 \sin ^{2} x=1 $
$\Rightarrow \sin ^{2} x=\frac{1}{4} $
$\Rightarrow \sin ^{2} x=\sin ^{2} \pi / 6$
$\Rightarrow x=n \pi \pm \pi / 6 $
this also gives $ 3$ solutions in $[0,3 \pi]$
Solutions of $16^{\sin ^{2} x}=16^{\cos ^{2} x}=10$ in $x \in[0,3 \pi]=6$.