## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{\sqrt{2}}{2}$
Divide both sides by $4$ to obtain: $\sin^{-1}{x} = \dfrac{\pi}{4}$ Recall: $y = \sin^{-1}{x} \hspace{15pt} \to \hspace{15pt} x = \sin{y}$ $\text{where } \hspace{15pt} -1 \leq x \leq 1 \hspace{15pt} \text{and} \hspace{15pt} -\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$ Thus, $\sin^{-1}{x}=\dfrac{\pi}{4} \rightarrow x = \sin{\dfrac{\pi}{4}}$ $x =\sin{\dfrac{\pi}{4}} = \boxed{\dfrac{\sqrt{2}}{2}}$