Question

If $e^{\cos x}-e^{-\cos x}=4$, then the value of $\cos x$is

• A. $\log (2+\sqrt{5})$
• B. $–\log (2+\sqrt{5})$
• C. $\log (–2+\sqrt{5})$
• D. None of the above

Answer 1

The correct option is D

None of the above

Finding the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{x}$:

It is given that:
$e^{\cos x}-e^{-\cos x}=4$

Put $e^{\cos x}=t$

$\begin{array}{rcl}t–\left(\frac{1}{t}\right)& =& 4\end{array}$

$\Rightarrow$$\begin{array}{rcl}{t}^2–4t–1& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}t& =& \frac{4\pm \sqrt{4^2}+4}{2}\\ & =& [2\pm \sqrt{2^2}+1]\\ & =& 2\pm \sqrt{5}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{e}^{\cos x}& =& 2+\sqrt{5}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\cos x& =& {\ln }_e(2+\sqrt{5})\end{array}$

$\Rightarrow$ $\begin{array}{rcl}e& =& 2.71\end{array}$

Hence, the correct answer is option D.