[Geometry]
Problem: Suppose $0^{\circ}<x<90^{\circ}$ and $2 \sin ^{2} x+\cos ^{2} x=\frac{25}{16}$. What is the value of $\sin x$ ?

Solution: Since $2 \sin ^{2} x+\cos ^{2} x=\frac{25}{16}$ and $\sin ^{2} x+\cos ^{2} x=1\left(\right.$ so $\left.\cos ^{2} x=1-\sin ^{2} x\right)$, then we get

$$
\begin{aligned}
2 \sin ^{2} x+\left(1-\sin ^{2} x\right) & =\frac{25}{16} \\
\sin ^{2} x & =\frac{25}{16}-1 \\
\sin ^{2} x & =\frac{9}{16} \\
\sin x & = \pm \frac{3}{4}
\end{aligned}
$$

so $\sin x=\frac{3}{4}$ since $\sin x>0$ because $0^{\circ}<x<90^{\circ}$.

Final answer: $\frac{3}{4}$