The solutions of the equation $ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $ are? I tried this:
$$ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $$
$$ 2\sin{2x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ 4\sin{x}\cos{x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ \sin{x}(1-\sin^2{x}) = \frac{2}{3\sqrt{3}} $$
Here, I tried to set $\sin x = t$
$$ t(1-t^2) = \frac{2}{3\sqrt{3}}, $$ but I don't know to resolve this.

HINT
To begin with, notice that $\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$. Hence it results that:
\begin{align*}
\sin(x) + \sin(3x) = \frac{8}{3\sqrt{3}} & \Longleftrightarrow 4\sin(x) - 4\sin^{3}(x) = \frac{8}{3\sqrt{3}}\\\\
& \Longleftrightarrow \sin(x) - \sin^{3}(x) = \frac{2\sqrt{3}}{9}
\end{align*}
By inspection, one concludes that
\begin{align*}
\sin(x) = \frac{1}{\sqrt{3}}
\end{align*}
satisfies the resulting equation.
From then on, you can factor the cubic equation to obtain a quadratic which is easy to deal with.
Can you take it from here?
