If $e^{sin x} - e^{- sin x} - 4 = 0$ then the number of real values of x is

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Solution

The correct option is D 0
let, $e^{sin x} = k$
$k - \frac{1}{k} - 4 = 0$
$k^{2} - 1 - 4 k = 0$
$\Rightarrow k = 2 \pm \sqrt{5}$
$e^{sin x} = k = 2 \pm \sqrt{5}$
$sin x = l n (2 \pm \sqrt{5}) > 1$
so, no solutions

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