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

You want to find $x$ s.t. $-(n+1) = \sum_{j=0}^{n} \sin (j\pi x)$. 
If $x\not = 1$, then the RHS is the imaginary part of
$$
X = \sum_{j=0}^n (\cos(j\pi x) + i\sin(j\pi x))= \sum_{j=0}^n \left( \cos(\pi x) + i\sin(\pi x) \right)^j = \frac{1 - \left( \cos(\pi x) + i\sin(\pi x) \right)^{n+1}}{1 - \left( \cos(\pi x) + i\sin(\pi x) \right)}
$$
$$
X = \frac{1 - (\cos \pi + i \sin \pi)^{x(n+1)}}{1 - (\cos \pi + i \sin \pi)^x} = \frac{1 - (-1)^{x(n+1)}}{1 - (-1)^x}
$$
Its imaginary part is of course $0$, so it doesn't depend on $x$. Just check what happens for $x = 1$.
