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

Answer: 9

Let $16^{{\sin }^{2}x}=y$
$16^{{\cos }^{2}x}=16^{1-{\sin }^{2}x}=16y^{-1}$
The equation can be written as $y+\frac{16}{y}=10$
$\Rightarrow y^2-10y+16=0$
$\Rightarrow (y-8)(y-2)=0$
$y=8,y=2$
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 $x\in [0,3\pi ]=6$.