Explanation

(a)
Given equation is
$2^{\cos{2x}} + 1 = {3.2}^{—\sin x}$
By taking option (a)
Let $x = n\pi$
When, $n = 1,\ \ x = \pi$
$\therefore\ \ \ 2^{\cos{2\pi}} + 1 = {3.2}^{- \sin\pi}$
$\Rightarrow \mathbf{\ \ }2 + 1 = 3.2{^\circ}\mathbf{\ \ } \Rightarrow \ \ 3 = 3$
When $n = 2,\ x = 2\pi$
$\therefore\ \ 2^{cos\ 4\pi} + 1 = 3.2^{- sin\ 2\pi}$
$\Rightarrow \ \ 2^{1} + 1 = 3.2{^\circ}$
$\Rightarrow \ \ 3 = 3$