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?
 A: Here's an approach that at least reduces the problem to solving a quartic polynomial.
Let $x=3u$ and $\phi=\theta+u$, so that the equation becomes $3\sin3\phi=2.5/\sin(\phi-u)$, or
$$6\sin3\phi(\sin\phi\cos u-\cos\phi\sin u)=5$$
If you can solve this for $\phi$ in terms of $u$, you can easily convert that solution to one for $\theta$ in terms of $x$.
Now $\sin3\phi=3\sin\phi-4\sin^3\phi$, hence, letting $s=\sin\phi$, we have
$$6(3s-4s^3)\left(s\cos u-\sqrt{1-s^2}\sin u\right)=5$$
or
$$6(3-4s^2)s^2\cos u-5=6(3-4s^2)\sqrt{s^2-s^4}\sin u$$
Squaring both sides produces a quartic polynomial in $s^2$, with coefficients that can be expressed in terms of $\cos u$ (since $\sin^2u=1-\cos^2u$).  I.e., letting $S=s^2=\sin^2\phi$ and $C=\cos u=\cos(x/3)$, we have
$$(6C(3-4S)S-5)^2=36(1-C^2)(3-4S)^2(S-S^2)$$
which expands out to a quartic in $S$. Because it's a quartic, this can, in principle, be solved for $S=s^2$, after which you can pick out solutions (if any) to the non-squared equation, but unless a miracle occurs (or there's a simplification I don't see), it looks like a mess. This leads me to wonder where this problem came from; it certainly doesn't strike me as a routine homework exercise.
A: 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 $$\tan(\frac{\theta}{2})=t$$
