How do I solve an equation with a variable both in a trigonometric function and outside of it?

Question

I realize that this should be fairly simple, and I'm not asking for the answer, but I'm curious as to concepts/equations I need to be familiar with to solve an equation of the form:

$\dfrac{\alpha}{\sin\alpha} = 1.05 $

How can I isolate $\alpha$?

Answer 1

This is a transcendental equation and you cannot solve this analytically. You can try graphical methods or numerical procedures to get an approximate solution to your problem.

Answer 2

This is too long for a comment.

Since SchrodingersCat already gave the answer to the question, I cannot resist the pleasure of showing once more what can be done using the superb approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ which was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (that is to say more than $1,400$ years ago) (see here).

Applied to the equation $$\dfrac{\alpha}{\sin(\alpha)} = k \qquad (k>1)$$ the problem reduces to $$4 \alpha ^2+ (16 k-4 \pi )\alpha-(16 k-5 \pi )\pi=0$$ the solution of which being $$ \alpha=\frac{\pi }{2}+\sqrt{4 k^2+2 \pi k-\pi ^2}-2 k$$ Using $k=1.05$, this approximation leads to $$\alpha \approx 0.5374$$ while the exact solution, given by Newton method for example, is $$\alpha \approx 0.5384$$