If sin−1(6x)+sin−1(63x)=−π2then the value of x is

We have sin−1(6x)+sin−1(63x)=−π2⇒sin-16x+π2-cos-163x=-π2 ⇒π-cos-163x=-sin-16x ⇒cos-1-63x=sin-1-6x Let  cos-1-63x=sin-1-6x=θ. ⇒cosθ=-63x and sinθ=-6x ∴sinθcosθ=-6x-63x ⇒tanθ=13 ⇒θ=π6 ∴sinθ=sinπ6=12=-6x ⇒x=-112