Explanation

(d)
Given equation is $e^{\sin x} - e^{- \sin x} - 4 = 0$
Let $e^{\sin x} = y$, then given equation can be written as
$y^{2} - 4y - 1 = 0 \Rightarrow y = 2 \pm \sqrt{5}$
But the value of $y = e^{\sin x}$ is always positive so we take only $y = 2 + \sqrt{5}$
$\Rightarrow \log_{e}{y = \log_{e}{(2 +}}\sqrt{5})$
$\Rightarrow \sin{x = \log_{e}{\left( 2 + \sqrt{5} \right) > 1}}$
Which is impossible since $\sin x$ cannot be greater than 1.
Hence, we cannot find any real value of $x$ which satisfies each given equation.