# Finding roots of Equation involving trig. functions.

In a problem of classical mechanics, I encounter the following equation:

$$\mu \sin^4 \theta + \cos \theta = 0 \qquad \mu > 0 \qquad \frac{\pi}{2} < \theta < \pi,$$

where $\mu$ is some constant containing things like mass $m$ and angular momentum $L$. I would like to find the roots in $\theta$ for it being in the interval $\left(\frac{\pi}{2},\pi \right)$, which is a condition I know for sure to be true. Mathematica gives me an extremely blown up solution for this equation. Is there any (nice) way to find the roots of this equation by hand?

Let $x=-\cos(\theta)$ so that

$$\mu(1-x^2)^2-x=0$$

This polynomial indeed has a single root in the range $[0,1]$ for all positive $\mu$.

We can rewrite the above equation as

$$\mu=\frac x{(1-x^2)^2}.$$

For small $x$, $x\approx\mu$.

For $x$ close to $1$, let $1-\epsilon$, we have

$$\mu=\frac{1-\epsilon}{(1-(1-\epsilon)^2)^2}\approx\frac1{4\epsilon^2}.$$

This gives us the approximation

$$x=1-\frac1{2\sqrt\mu}.$$

You can refine the root numerically in the range

$$[\mu,1-\frac1{2\sqrt\mu}].$$

For a "manual" method, you can plot the relation $\mu=f(x)$ as accurately as possible. Then for a given value of $\mu$, find the corresponding $x$ on the plot and use it for a starting value of Newton's iterations.

Update:

The $\mu$ curve has a vertical asymptote at $x=1$, which makes it more difficult to handle. We can discard it by considering the function

$$\frac\mu{\mu+1}=\frac x{(1-x^2)^2\left(\frac x{(1-x^2)^2}+1\right)}=\frac x{(1-x^2)^2+x}.$$

It turns out that the function is fairly well approximated by $x$ in the range of interest, so that a good initial approximation is simply

$$x=\frac\mu{\mu+1} !$$

• For $\mu>1$ the given range for numerical refinement becomes useless, right? – Matt L. Dec 16 '15 at 21:53
• @MattL.You are right. – Yves Daoust Dec 16 '15 at 21:57

You can avoid ugly analytical formulas but you'll need to use a numerical method to solve for the roots of a polynomial.

$$\mu\sin^4\theta-\sqrt{1-\sin^2\theta}=0$$

(because $\pi/2<\theta<\pi$), from which you get

$$\mu^2\sin^8\theta+\sin^2\theta - 1=0$$

With $x=\sin\theta$ this is equivalent to

$$x^8+\frac{1}{\mu^2}x^2-\frac{1}{\mu^2}=0\tag{1}$$

Solve for the roots of $(1)$, choose the one which is real-valued and positive, and compute the desired angle as

$$\theta_0=\pi-\arcsin x_0$$

(again, because $\pi/2<\theta<\pi$).

• Shouldn't your first equation read $\mu\sin^4\theta+\sqrt{1-\sin^2\theta}=0$ instead of $\mu\sin^4\theta-\sqrt{1-\sin^2\theta}=0$ ? – K. Rmth Dec 16 '15 at 21:35
• @K.Rmth: No, because $\cos\theta <0$ for the given interval of $\theta$. – Matt L. Dec 16 '15 at 21:37
• Understood. :-) – K. Rmth Dec 16 '15 at 21:39