Which of the following is the solution set of the equation `sin^-1 x = cos ^-1 x+ sin ^-1(3x-2)`

Options

A. `[0,1/3]`
B. `[1/3,2/3]`
C. `[0,2/3]`
D. none of these

Explanation

For the existance of the given equation we must have
`-1lexle1and-1le3x-1le1rarr0lexle2/3`
Now
`sin^(1)x=cos^(-1)x+sin^(-1)(3x-1)`
`rarr sin^(-1)x-cos^(-1)x=sin^(-1)(3x-1)`
`rarr 2sin^(-1)x-(pi)/(2)=sin^(-1)(3x-1)`
`rarr sin(2sin^(-1)x-(pi)/(2))=sin(sin^(-1))(3x-1)`
`rarr =cos(2sin^(-1)x)=3x-1`
`rarr -{1-2sin^(2)(sin^(-1)x))}=3x-1`
`rarr -(1-2x^(2)=3x-1)`
`rarr 2x^(2)-3x=0rarr x=0 3/2rarrx=0`