How to express $\theta$ in terms of $x$ where $3\sin(3\theta+x)=\frac{2.5}{\sin\theta}$?

Question

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?

Answer 1

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 $$\tan(\frac{\theta}{2})=t$$

Answer 2

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.