<s>[INST] Finding all real $x$ such that $1+\sum_{j=1}^n\sin{\frac{j\pi x}{n+1}} = 0$, where $n=18$. 
The task is to find all $ x \in \mathbb R $ such that
$$ 1 + \sum_{j=1}^n \sin{\frac{j\pi x}{n + 1}} = 0, \qquad n = 18 $$

What I have tried
Using the following formulas:
$$1. \sin{x} + \sin{y} = 2\sin{\frac{x + y}{2}}\cos{\frac{x - y}{2}} $$
$$2. \cos{x} + \cos{y} = 2\cos{\frac{x + y}{2}}\cos{\frac{x - y}{2}} $$
$$3. \sin{x}\sin{y} = \frac{1}{2} (\cos{(x - y)} - \cos{(x + y)}) $$
First approach
$$ 1 + \sin{\frac{\pi x}{19}} + \sin{\frac{18\pi x}{19}} + \sin{\frac{2 \pi x}{19}} + \sin{\frac{17\pi x}{19}} + ... + \sin{\frac{9 \pi x}{19}} + \sin{\frac{10 \pi x}{19}} = 0 $$
Using formula 1:
$$ 1 + 2\sin{\frac{\pi x}{2}} (\cos{\frac{17\pi x}{19}} + \cos{\frac{15\pi x}{19}} + ... + \cos{\frac{\pi x}{19}})$$
Skipping some steps, we get:
$$ 1 + 2\sin{\frac{\pi x}{2}}\cos{\frac{9\pi x}{38}}(2\cos{\frac{3\pi x}{38} + 1})(2\cos{\frac{\pi x}{38}} + 1) = 0 $$
From here I couldn't find a way to go further
Second approach
Multiply both sides by $ \sin{\frac{\pi x}{38}} $:
$$ \sin{\frac{\pi x}{38}} (1 + \sin{\frac{\pi x}{19}} + \sin{\frac{2\pi x}{19}} + \sin{\frac{3 \pi x}{19}} + ... + \sin{\frac{17 \pi x}{19}} + \sin{\frac{18 \pi x}{19}}) = 0 $$
Using formula number 3
$$ \sin{\frac{\pi x}{38}} + \frac{1}{2}(\cos{\frac{\pi x}{38}} - \cos{\frac{3\pi x}{38}} + \cos{\frac{3\pi x}{38}} - \cos{\frac{5\pi x}{38}} + ... + \cos{\frac{35\pi x}{38}} - \cos{\frac{37\pi x}{38}}) = 0$$
$$ \sin{\frac{\pi x}{38}} - \sin{\frac{\pi x}{2}}\sin{\frac{18\pi x}{38}} = 0 $$
From here I couldn't find any way to go further
Any hints will be appreciated!
EDIT:
The upper-most formula had a mistake, but now it is up to date.
PS
The problem is supposed to be solvable assuming you have high-school level math (i.e only simple trigonometric formulas are allowed, no complex numbers, no derivatives for the sine and cosine functions).
Also, this equation on Wolfram Alpha shows the following solution:
$ x = \frac{19}{2}(4n - 1), n \in \mathbb Z $
Update 2: Using @Doug M's answer, I managed to conclude that ($ \sin{\frac{\pi x}{19}} = 1 $ and $ \cos{\frac{\pi x}{19}} = 0 $), which equalates to $ \frac{\pi x}{19} = \frac{\pi}{2} + 2\pi n, n \in \mathbb Z $ is one of the solutions to the equation
 [/INST] Multiply the series by $\frac {\sin \frac {\pi x}{19}}{\sin \frac {\pi x}{19}}$
Then use 
$\sin A\sin B = \frac 12 \cos  (A-B) - \frac 12 \cos (A+B)$
And the series will telescope.
$1+\sum_\limits{j=1}^{18} \frac {\sin \frac{j\pi}{19}\sin \frac {\pi x}{19}}{\sin\frac {\pi x}{19}}$
$1+\sum_\limits{j=1}^{18} \left(\frac {\cos \frac{(j-1)\pi}{19}}{2\sin\frac {\pi x}{19}} - \frac {\cos \frac{(j+1)\pi}{19}}{2\sin\frac {\pi x}{19}}\right)$
$1+\frac{\cos 0 + \cos \frac {\pi x}{19} - \cos \frac {18\pi x}{19} - \cos \pi x}{2\sin \frac {\pi x}{19}} = 0$
let $y = \frac {\pi x}{19}$
$2\sin y + 1 + \cos y - \cos 18y - \cos 19y = 0$
(multiplying through by $\sin y$ has introduced an artificial solution at $\sin y = 0$)
$y = \frac {3\pi}{2} + 2k\pi$ is a solution, implying $x = \frac {57}{2}$
As for the others....
If we expand the $\cos 19 y, \cos 18 y$ terms we could turn this into a function of powers of $cos y$ and $sin y$ but we still have a 19 (or 17 degree when we factor out the two identified factors) degree polynomial to factor.
I put it into wolfram alpha and none of the other solutions are particularly nice.
https://www.wolframalpha.com/input/?i=sin+y++%2B+1+%2B+cos+y+-+cos+18y+-+cos+19y+%3D+0
This is as far as I know how to take it.
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