Question

The value of $\mathrm{x}$ which satisfies $8^{1+\mathrm{cosx}+{\cos }^2\mathrm{x}+...}=64$ in $[-\mathrm{\pi },\mathrm{\pi }]$ is

• A. $\pm \frac{\mathrm{\pi }}{2},\pm \frac{\mathrm{\pi }}{3}$
• B. $\pm \frac{\mathrm{\pi }}{3}$
• C. $\pm \frac{\mathrm{\pi }}{2},\pm \frac{\mathrm{\pi }}{6}$
• D. $\pm \frac{\mathrm{\pi }}{6},\pm \frac{\mathrm{\pi }}{3}$

Answer 1

The correct option is B

$\pm \frac{\mathrm{\pi }}{3}$

Explanation for the correct option:

Given: $8^{1+\mathrm{cosx}+{\cos }^2\mathrm{x}+...}=64$

$1+\cos x+{\cos }^{2}x+.....$ is an infinite geometric progression with $\mathrm{a}=1,\mathrm{r}=\mathrm{cosx}$

The sum of infinite GP is $\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{1}{1-\mathrm{cosx}}$

$\begin{array}{crcl}\Rightarrow & 8^{\frac{1}{1-\cos x}}& =& 8^2\\ \Rightarrow & \frac{1}{1-\cos x}& =& 2\\ \Rightarrow & 1& =& 2(1-\cos x)\\ \Rightarrow & \frac{1}{2}& =& 1-\cos x\\ \Rightarrow & \cos x& =& 1-\frac{1}{2}\\ \Rightarrow & \cos x& =& \frac{1}{2}\\ \Rightarrow & \cos x& =& \cos \frac{\mathrm{\pi }}{3}\\ \Rightarrow & x& =& \pm \frac{\mathrm{\pi }}{3}\end{array}$

Hence option B is the correct.