[Algebra]
Problem: Suppose that $f(a)=2 a^{2}-3 a+1$ for all real numbers $a$ and $g(b)=\log _{\frac{1}{2}} b$ for all $b>0$. Determine all $\theta$ with $0 \leq \theta \leq 2 \pi$ for which $f(g(\sin \theta))=0$.

Solution: Using the definition of $f$, the following equations are equivalent:

$$
\begin{aligned}
f(a) & =0 \\
2 a^{2}-3 a+1 & =0 \\
(a-1)(2 a-1) & =0
\end{aligned}
$$

Therefore, $f(a)=0$ exactly when $a=1$ or $a=\frac{1}{2}$.

Thus, $f(g(\sin \theta))=0$ exactly when $g(\sin \theta)=1$ or $g(\sin \theta)=\frac{1}{2}$.

Using the definition of $g$,

- $g(b)=1$ exactly when $\log _{\frac{1}{2}} b=1$, which gives $b=\left(\frac{1}{2}\right)^{1}=\frac{1}{2}$, and
- $g(b)=1 / 2$ exactly when $\log _{\frac{1}{2}} b=1 / 2$, which gives $b=\left(\frac{1}{2}\right)^{1 / 2}=\frac{1}{\sqrt{2}}$.

Therefore, $f(g(\sin \theta))=0$ exactly when $\sin \theta=\frac{1}{2}$ or $\sin \theta=\frac{1}{\sqrt{2}}$.

Since $0 \leq \theta \leq 2 \pi$, the solutions are $\theta=\frac{1}{6} \pi, \frac{5}{6} \pi, \frac{1}{4} \pi, \frac{3}{4} \pi$.

Final answer: $\frac{1}{6} \pi, \frac{5}{6} \pi, \frac{1}{4} \pi, \frac{3}{4} \pi$