Q. 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 these
Explanation
(a)
Let $e^{\cos x} = y$
Then, $y - \frac{1}{y} = 4\ \ \ \Rightarrow \ \ y^{2} - 4y - 1 = 0$
$\Rightarrow \ \ y = \frac{- ( - 4) \pm \sqrt{16 - 4 \times ( - 1)}}{2} \Rightarrow y = \frac{4 \pm 2\sqrt{5}}{2}$
$\Rightarrow \ \ \ y = 2 + \sqrt{5} = e^{\cos x}$ [$\because$ exponential is always positive]
$\Rightarrow \ \cos{x = \log{(2 + \sqrt{5})}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.