The solution of `sin^(-1)x-sin^(-1)2x=pm(pi)/(3)` is

Options

A. `pm(1)/(3)`
B. `pm(1)/(4)`
C. `pm(sqrt(3))/(2)`
D. `pm(1)/(2)`

Answer

D

Explanation

`sin^(-1)x-sin^(-1)2x=pm (pi)/(3)`
`rArr sin^(-1)x-sin^(-1)(pm(sqrt(3))/(2))=sin^(-1)2x`
`rArr sin^(-1)[x sqrt(1-(3)/(4))-(pm (sqrt(3))/(2)sqrt(1-x^(2)))]=sin^(-1)2x`
`rArr (x)/(2)-(pm(sqrt(3))/(2)sqrt(1-x^(2)))=2x`
`rArr -(pm sqrt(3)sqrt(1-x^(2)))=3x`
On squaring both sides, we get
`3(1-x^(2))=9x^(2)`
`rArr 4x^(2)=1`
`rArr x = pm (1)/(2)`