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.

As you obtained,
$ \displaystyle t-t^3 = \frac{2}{3\sqrt{3}}~, \text {where } t = \sin x$
$ \displaystyle t - t^3 = \frac{1}{\sqrt3} - \frac{1}{3 \sqrt3} \implies \left(\frac{1}{\sqrt3}\right)^3 - t^3 = \left(\frac{1}{\sqrt3} - t\right)$
Now using the fact that $a^3 - b^3 = (a-b)(a^2 + b^2 + ab)$
One of the obvious solution is $~t = \dfrac{1}{\sqrt3}$.
If $~t \ne \dfrac{1}{\sqrt3}, $ we have
$ ~\displaystyle t^2 + \frac{1}{3} + \frac{t}{\sqrt3} = 1$
i.e. $~ \displaystyle \left(t + \frac{1}{2 \sqrt3}\right)^2 = \frac{3}{4}~$. That gives $ \displaystyle t = \frac{1}{\sqrt3}, - \frac{2}{\sqrt3}$
As $~- \dfrac{2}{\sqrt3} \lt - 1$, that is not a valid solution.
So the only solution is $\sin x = \dfrac{1}{\sqrt3}$.
