Question

Solve the following:

`sin^-1x+sin^-1  2x=pi/3`

Answer 1

We know

`sin^-1x+sin^-1y=sin^-1[xsqrt(1-y^2)+ysqrt(1-x^2)]`

∴ `sin^-1x+sin^-1  2x=pi/3`

⇒ `sin^-1x+sin^-1  2x=sin^-1(sqrt3/2)`

⇒ `sin^-1x-sin^-1(sqrt3/2)=-sin^-1  2x`

⇒ `sin^-1[xsqrt(1-3/4)+sqrt3/2sqrt(1-x^2)]=-sin^-1  2x`

⇒ `sin^-1[x/2+sqrt3/2sqrt(1-x^2)]=sin^-1(-2x)`

⇒ `x/2+sqrt3/2sqrt(1-x^2)=-2x`

⇒ `x+sqrt3sqrt(1-x^2)=-4x`

⇒ `5x=-sqrt3sqrt(1-x^2)`

Squaring both the sides,

`25x^2=3-3x^2`

⇒ `28x^2=3`

⇒ `x=+-1/2sqrt(3/7)`