Question

Solve the following equation for x:

`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`

Answer 1

`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`

`=>tan^-1((2sinx)/(1-sin^2x))=tan^-1(2sinx)`     `[because2tan^-1x=tan^-1((2x)/(1-x^2))]`

`=>(2sinx)/(1-sin^2x)=2sinx`

`=>2sinx=2sinx-2sin^3x`

`=>2sin^2x=0`

`=>sinx=0`

`=>x=0`