Question

If `(sin^-1x)^2+(cos^-1x)^2=(17pi^2)/36,`  Find x

Answer 1

`(sin^-1x)^2+(cos^-1x)^2=(17pi^2)/36`

 ⇒ `(sin^-1x)^2+(pi/2-sin^-1x)^2=(17pi^2)/36`

Let `sin^-1x=y`

`therefore(y)^2+(pi/2-y)^2=(17pi^2)/36`

⇒ `y^2+pi^2/4+y^2-2xxpi/2xxy=(17pi^2)/36`

⇒ `2y^2-piy=(2pi^2)/9`

⇒ `18y^2-9piy-2pi^2=0`

⇒ `18y^2-12piy+3piy-2pi^2=0`

⇒ `6y(3y-2pi)+pi(3y+2pi)=0`

⇒ `(3y-2pi)(6y+pi)=0`

⇒ `y=pi/6`   [Neglecting `y=2/3pi` as it is not satisfying the question]

`thereforex=siny=sin(-pi/6)=-1/2`