Trig Challenge: Solutions to $\sin a \sin (2a) \sin (3a)$?

[anemone]

How many solutions does the equation

$\sin a \sin (2a) \sin (3a) \cdots \sin (11a) \sin (12a) =0$

have in the interval $(0,\,\pi]$?

 

[Nono713]

Spoiler

It is clear that $a = \pi p/q$ for every fraction of the form $p / q$ with $1 \leq q \leq 12$, $0 < p \leq q$ is a solution to some product of the expression and hence a solution to the equation. Furthermore all solutions are of this form since $p/q \leq 1$. Hence the number of solutions to the expression is given by the length of the Farey sequence $F_{12}$ minus 1 (since it includes zero and we exclude it). Now we have:
$$\lvert F_1 \rvert = 2 ~ ~ ~ \text{and} ~ ~ ~ \lvert F_n \rvert = \lvert F_{n - 1} \rvert + \varphi(n)$$
Which gives us the general expression:
$$\lvert F_n \rvert = 1 + \sum_{m = 1}^n \varphi(m)$$
Therefore:
$$\lvert F_{12} \rvert = 1 + \varphi(1) + \varphi(2) + \cdots + \varphi(12)$$
Which gives:
$$\lvert F_{12} \rvert = 1 + 1 + 1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 + 4 + 10 + 4 = 47$$
And therefore we conclude there are $46$ solutions to the original expression.

 

[anemone]

Well done, Bacterius! And thanks for participating!

The solution that I have is more or less the same as yours.:)