How to express $\theta$ in terms of $x$ where $3\sin(3\theta+x)=\frac{2.5}{\sin\theta}$? I tried to solve it by using compound angle formulas but in the end I could not leave $\theta$ alone.
It goes like this:
\begin{align}
& \frac{2.5}{3}=\sin(3\theta+x)\sin\theta \\[10pt]
& \sin(3\theta+x)=\sin(2\theta+\theta)\cos(x)+\sin(x)\cos(2\theta+\theta) \\[10pt]
= {} & [(\sin(2\theta)\cos(\theta)+\sin(\theta)\cos(2\theta))]\cos(x)+[(\cos(2\theta)\cos(\theta)-\sin(2\theta)\sin(\theta))]\sin(x)
\end{align}
Then I did a couple more steps but I couldn't solve it this way, is there any other way to solve it algebraically?

your equation is equivalent to $$4\,\cos \left( x \right)  \left( \sin \left( \theta \right)  \right) ^
{2} \left( \cos \left( \theta \right)  \right) ^{2}-\cos \left( x
 \right)  \left( \sin \left( \theta \right)  \right) ^{2}+4\,\sin
 \left( \theta \right) \sin \left( x \right)  \left( \cos \left( 
\theta \right)  \right) ^{3}-3\,\sin \left( \theta \right) \sin
 \left( x \right) \cos \left( \theta \right) 
=2.5$$
you can Substitute
$$\sin(\theta)=\frac{2\tan(\frac{\theta}{2})}{1+\tan(\frac{\theta}{2})^2}$$
and $$\cos(\theta)=\frac{1-\tan(\frac{\theta}{2})^2}{1+\tan(\frac{\theta}{2})^2}$$
and then you can Substitute $$\tan(\frac{\theta}{2})=t$$
