Question

If $\ln {{e}^{2{{\cos }^{2}}x-1}}=\cos x,0 < x\le \pi $, then the value of x is
[a] 0
[b] $\dfrac{\pi }{6}$
[c] $\dfrac{\pi }{3}$
[d] $\dfrac{2\pi }{3}$

Answer 1

Hint: Use the fact that ${{\log }_{a}}{{a}^{x}}=x$. Hence prove that the given equation is equivalent to the equation $2{{\cos }^{2}}x-\cos x-1=0$. Put $\cos x=t$ and use the fact that the solution of the quadratic equation $a{{x}^{2}}+bx+c=0$ is given by$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ or use the method of splitting the middle term or completing the square. Hence determine the value of $\cos x$. Use the fact that $-1\le \cos x\le 1$ to remove the extraneous roots (if any) and hence find the value of cosx. Use the fact that if $\cos x=\cos y$, then $x=2n\pi \pm y,n\in \mathbb{Z}$. Use the fact that $0 < x\le \pi $ and hence find the value of n and hence find the value of x.

Complete step-by-step answer:
We have
$\ln {{e}^{2{{\cos }^{2}}x-1}}=\cos x$
We know that ${{\log }_{a}}{{a}^{x}}=x$
Hence, we have
$\ln {{e}^{2{{\cos }^{2}}x-1}}=2{{\cos }^{2}}x-1$
Substituting the value of $\ln {{e}^{2{{\cos }^{2}}x-1}}$ in the original equation, we get
$2{{\cos }^{2}}x-1=\cos x$
Subtracting cosx from both sides, we get
$2{{\cos }^{2}}x-\cos x-1=0$
Put cosx = t, we get
$2{{t}^{2}}-t-1=0$
We use method of splitting the middle term to solve the quadratic equation
We have $1=2-1,2=2\times 1$
Hence, we have
$2{{t}^{2}}-2t+t-1=0$
Taking 2t common from the first two terms and 1 common from the last two terms, we get
$2t\left( t-1 \right)+1\left( t-1 \right)=0$
Taking t-1 common from these two terms, we get
$\left( 2t+1 \right)\left( t-1 \right)=0$
Hence by zero product property, we have
$\begin{align}
  & 2t+1=0\text{ or }t-1=0 \\
 & \Rightarrow t=\dfrac{-1}{2},1 \\
\end{align}$
Since $0 < x\le \pi $, we have $\cos x=t\ne 1$
Hence, we have
$t=-\dfrac{1}{2}$
Reverting to original variable, we get
$\cos x=-\dfrac{1}{2}=\cos \left( \dfrac{2\pi }{3} \right)$
We know that $\cos x=\cos y$, then $x=2n\pi \pm y,n\in \mathbb{Z}$
Hence, we have
$x=2n\pi \pm \dfrac{2\pi }{3},n\in \mathbb{Z}$
We have
$0 < x\le \pi $
Hence, we have
 $\begin{align}
  & 2n\pi \pm \dfrac{2\pi }{3} > 0 \\
 & \Rightarrow 2n\pi > \pm \dfrac{2\pi }{3} \\
 & \Rightarrow n > \pm \dfrac{1}{3} \\
 & \Rightarrow n\ge 0 \\
\end{align}$
Also, we have
$\begin{align}
  & 2n\pi \pm \dfrac{2\pi }{3}\le \pi \\
 & \Rightarrow 2n\pi \le \pi +\dfrac{2\pi }{3}\ \ or\ \ 2n\pi \le \pi -\dfrac{2\pi }{3} \\
 & \Rightarrow n\le 0\ \ or\ n\le -1 \\
\end{align}$
Hence, we have n = 0.
Hence, we have $x=\dfrac{2\pi }{3}$

So, the correct answer is “Option d”.

Note: [1] Verification:
We can verify the correctness of our solution by checking that $x=\dfrac{2\pi }{3}$ satisfies the equation $\ln {{e}^{2{{\cos }^{2}}x-1}}=\cos x$
We have $2{{\cos }^{2}}\left( \dfrac{2\pi }{3} \right)-1=2{{\left( \dfrac{-1}{2} \right)}^{2}}-1=\dfrac{-1}{2}$
Also, we have $\cos \left( \dfrac{2\pi }{3} \right)=-\dfrac{1}{2}$
Hence, we have $\ln {{e}^{-\dfrac{1}{2}}}=-\dfrac{1}{2}$
We know that if $\ln x=a$, then $x={{e}^{a}}$
Hence, we have
${{e}^{-\dfrac{1}{2}}}={{e}^{-\dfrac{1}{2}}}$ which is true.
Hence $x=\dfrac{2\pi }{3}$ satisfies our equation and hence our solution is correct.
Hence option [d] is correct.