trigonometric equation opening Solve: 

$$ \sin x + \sin 3x + \sin 5x  = 0 . $$ 

Attempt at a solution: applying formulas for summation of sine we get after a series of operations: $ \sin x(8 \cos x \cos 2x \cos x + 1) = 0$ equaling sine to $0$ we get one solution $180k$. comparing the other factor we eventually get $\cos^2 x = 0.25 $ this in turn produces the solutions plus/minus $120 + 180 k$ and plus/minus $60 + 180 k$ , giving us $5$ solutions in total.  ( $K \in Z $ )
Is this ok? 
 A: Yes, expressed in degrees, you've covered all the bases. 
(Nit pick: why not specify that $k \in \mathbb Z$?)
Nice work on solving the equation.
A: Using $e^{xi}=\cos x+i\sin x$, we can easily derive a canonical form for $\sin(ax)$, for $a\in\mathbb{N}$.
For example, $e^{3xi}=(\cos x+i\sin x)^3=\cos 3x+i\sin 3x$.  Equating the imaginary parts of each expression, we obtain $$3\cos^2x \sin x-\sin^3x=\sin 3x$$  If we reuse this technique for 5, $e^{5xi}=(\cos x+i\sin x)^5=\cos 5x+i\sin 5x$.  Equating the imaginary parts again, 
$$ 5\cos^4x\sin x-10\cos^2x\sin^3x+\sin^5x=\sin 5x $$
So, your question reduces to
$$ \begin{align}
\sin x+3\cos^2x \sin x-\sin^3x+5\cos^4x\sin x-10\cos^2x\sin^3x+\sin^5x&=0 \\
\sin x \left(1+3(1-\sin^2x)-\sin^2x+5(1-\sin^2x)^2-10\sin^2x(1-\sin^2x)+\sin^4x\right)&=0 \\
\sin x(9-24\sin^2x+16\sin^4x)&=0
\end{align}$$
Now, we have $\sin x=0$ or $9-24\sin^2x+16\sin^4x=0$.  Using the quadratic formula,
$$ \sin^2x=\frac{24\pm\sqrt{24^2-4\cdot9\cdot16}}{32}=\frac{3}{4} $$
So your solutions are $\sin x=0,\ \pm\frac{\sqrt{3}}{2}$, or when $x$ is an integer multiple of $\frac{\pi}{3}$
A: We consider in general the following sum:
$$
\sum_{k=0}^{n-1}\sin(2k+1)x
$$
Using the formula $\sin x = \displaystyle\frac{1}{2i}(e^{ix}-e^{-ix})$, we can simplify this sum:
$$
\frac{1}{2i}\left(\sum_{k=0}^{n-1} e^{ix(2k+1)}-e^{-ix(2k+1)}\right)=\frac{1}{2i}\left(\sum_{k=0}^{n-1} \alpha\cdot \beta^k-\alpha^{-1}\beta^{-k}\right)
$$
Where $\alpha=e^{ix}$ and $\beta = \alpha^2=e^{2ix}$. Using geometric series this simplifies to 
$$
\frac{1}{2i}\left(\alpha\left(\frac{\beta^n-1}{\beta-1}\right)-\alpha^{-1}\left(\frac{\beta^{-n}-1}{\beta^{-1}-1}\right)\right)=\frac{1}{2i}\left(\frac{\alpha^{2n}+\alpha^{-2n}-2}{\alpha-\alpha^{-1}}\right)
$$
But note that $\alpha-\alpha^{-1}=2i\sin x$, and that $$\alpha^{2n}+\alpha^{-2n}-2=(\alpha^{n}-\alpha^{-n})^2=(2i\sin{nx})^2$$
So the last expression simplifies to
$$
\frac{(2i)^2\sin^2{nx}}{(2i)^2\sin{x}}=\csc x\sin^2{nx}
$$
Applying this to the given problem, it reduces to solving
$$
\csc x\sin^2{3x}=0
$$
But since $\csc x\neq 0$, we need $\sin{3x}=0$ so we get $x=\frac{\pi}{3}n$ for integer $n$. The only possible exception occurs when $x$ is a multiple of $\pi$, which would give $\csc x$ infinite, but we can easily check that $x$ satisfies the original equation in these cases as well.
