The number of real roots of the equation {e^{ | vedclass

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Question

(d) Given equation ${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$

Let ${e^{\sin x}} = y$, then given equation can be written as

${y^2} - 4y - 1 = 0$==

Vedclass · Accepted Answer

None

Vedclass · Answer

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

The number of real roots of the equation ${e^{\sin x}} - {e^{ - \sin x}} - 4$ $ = 0$ are

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