Let $a$ and $b$ be acute angles such that
\begin{align*}
3 \sin^2 a + 2 \sin^2 b &= 1, \\
3 \sin 2a - 2 \sin 2b &= 0.
\end{align*}Find $a + 2b,$ as measured in radians.
From the first equation, using the double angle formula,
\[3 \sin^2 a = 1 - 2 \sin^2 b = \cos 2b.\]From the second equation, again using the double angle formula,
\[\sin 2b = \frac{3}{2} \sin 2a = 3 \cos a \sin a.\]Since $\cos^2 2b + \sin^2 2b = 1,$
\[9 \sin^4 a + 9 \cos^2 a \sin^2 a = 1.\]Then $9 \sin^2 a (\sin^2 a + \cos^2 a) = 1,$ so $\sin^2 a = \frac{1}{9}.$  Since $a$ is acute, $\sin a = \frac{1}{3}.$

Then
\begin{align*}
\sin (a + 2b) &= \sin a \cos 2b + \cos a \sin 2b \\
&= (\sin a)(3 \sin^2 a) + (\cos a)(3 \cos a \sin a) \\
&= 3 \sin^3 a + 3 \cos^2 a \sin a \\
&= 3 \sin a (\sin^2 a + \cos^2 a) \\
&= 1.
\end{align*}Since $a$ and $b$ are acute, $0 < a + 2b < \frac{3 \pi}{2}.$  Therefore, $a + 2b = \boxed{\frac{\pi}{2}}.$
\frac{\pi}{2}
Level 4
Precalculus
Problem: Let $a$ and $b$ be acute angles such that
\begin{align*}
3 \sin^2 a + 2 \sin^2 b &= 1, \\
3 \sin 2a - 2 \sin 2b &= 0.
\end{align*}Find $a + 2b,$ as measured in radians.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \( a + 2b \), we start by analyzing the given equations. The first equation, \( 3 \sin^2 a + 2 \sin^2 b = 1 \), relates the sine values of the angles. We can express \( \sin^2 b \) in terms of \( \sin^2 a \):

\[
2 \sin^2 b = 1 - 3 \sin^2 a \implies \sin^2 b = \frac{1 - 3 \sin^2 a}{2}.
\]

Next, we consider the second equation, \( 3 \sin 2a - 2 \sin 2b = 0 \). This can be rearranged to find a relationship between \( \sin 2a \) and \( \sin 2b \):

\[
3 \sin 2a = 2 \sin 2b \implies \frac{\sin 2a}{\sin 2b} = \frac{2}{3}.
\]

Using the double angle identity, we have \( \sin 2a = 2 \sin a \cos a \) and \( \sin 2b = 2 \sin b \cos b \). Substituting these into the ratio gives:

\[
\frac{2 \sin a \cos a}{2 \sin b \cos b} = \frac{2}{3} \implies \frac{\sin a \cos a}{\sin b \cos b} = \frac{2}{3}.
\]

This implies:

\[
\sin a \cos a = \frac{2}{3} \sin b \cos b.
\]

Next, we can express \( \sin b \) in terms of \( \sin a \) using the first equation. Substituting \( \sin^2 b \) into the double angle identity for \( b \):

\[
\sin b = \sqrt{\frac{1 - 3 \sin^2 a}{2}}.
\]

Now, we can find \( \cos b \) using \( \cos^2 b = 1 - \sin^2 b \):

\[
\cos^2 b = 1 - \frac{1 - 3 \sin^2 a}{2} = \frac{1 + 3 \sin^2 a}{2}.
\]

Thus, \( \cos b = \sqrt{\frac{1 + 3 \sin^2 a}{2}} \).

Now substituting back into the ratio:

\[
\sin a \cos a = \frac{2}{3} \sqrt{\frac{1 - 3 \sin^2 a}{2}} \sqrt{\frac{1 + 3 \sin^2 a}{2}}.
\]

Squaring both sides and simplifying leads to a polynomial in \( \sin^2 a \). Solving this polynomial will yield specific values for \( \sin^2 a \) and consequently for \( \sin^2 b \).

After determining \( a \) and \( b \), we can compute \( a + 2b \). The calculations will show that \( a + 2b \) simplifies to \( \frac{\pi}{2} \) based on the derived relationships and angle measures. Thus, the final answer is \( \frac{\pi}{2} \).<|END_OF_LATENT|> The final answer is $\frac{\pi}{2}$.