Answer
$\arccos \dfrac{x}{6}=\arcsin \dfrac{\sqrt {36-x^2}}{6}$
Work Step by Step
Consider $\theta=\arcsin \dfrac{\sqrt {36-x^2}}{6}$
We can see that $\sqrt {36-x^2}$ is a positive quanity and $\theta$ must lie in the first quadrant.
So, we have: $\cos \theta=\dfrac{x}{6} $
and $\theta =\arccos \dfrac{x}{6} $
and $\arccos \dfrac{x}{6}=\arcsin \dfrac{\sqrt {36-x^2}}{6}$
