Question

If, $x∈(0,π)$ satisfies the equation $6^{1+sinx+sin^2x....}=36$ ,then the value of $x$ is_____.

• A. $0$
• B. $\dfrac{\pi}{3}$
• C. $\dfrac{\pi}{6}$
• D. $\dfrac{\pi}{4}$
• E. $\dfrac{\pi}{2}$

Answer 1

Answer: (C)

$\dfrac{\pi}{6}$

Answer 2

Given that:

$x∈(0,π)$ satisfies the equation $6^{1+sinx+sin^2x......}=36$

Then,

$6^{1+sinx+sin^2x......}=36$

$⇒6^{1+sinx+sin^2x......}=6^2$

$⇒1+sinx+sin^2x......=2$

This represents an infinite G.P series where we can write , first term $a =sinx$ and common ratio

$r= sin^2(x)$

The sum of an infinite geometric series is given by the formula

$S= \dfrac{a}{1-r}$

by substituting values we get

$2=\dfrac{sinx}{1-sin^{2}x}$

$⇒2-2sin^{2}x-sinx=0$

$⇒2sin^{2}x+sinx-2=0$

$⇒(2sinx-1)(sinx+2)=0$

Therefore on solving the above expression we get

$x=\dfrac{\pi}{6}$ (_Ans.)