Question

$2^{\cos 2x}={3.2}^{{\cos }^{2}x}-4$

Answer 1

$2^{2c^2-1}={3.2}^{c^2}-4$. Put $2^{c^2}=y$
or $y^2=2(3y-4)$
or $y^2-6y+8=0$
$\therefore (y-4)(y-2)=0$
$\therefore y=2c^2=2^2$ or $2^1$
$\therefore {\cos }^{2}x=2$ or ${\cos }^{2}x=1$
$\therefore {\sin }^{2}x=0$ or $x=n\pi$.