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

The Correct Option is C

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.)