Answer
$\dfrac{\pi}{3}$
Work Step by Step
Suppose $\theta =\sin (\dfrac{\pi}{3})$
This gives: $\sin^{-1} \theta=\sin^{-1}[\sin(0.9)]$
The inverse property states that $\sin^{-1}(\sin a)=a $ satisfies for every $ a $ in the range $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$ .
Therefore, we have $ \sin^{-1}[\sin(\dfrac{\pi}{3})]=\dfrac{\pi}{3}$
