If $\sin x +\sin 2x + \sin 3x = \sin y\:$ and $\:\cos x + \cos 2x + \cos 3x =\cos y$, If  $\sin x +\sin 2x + \sin 3x = \sin y\:$ and $\:\cos x + \cos 2x + \cos 3x =\cos y$, then $x$ is equal to
(a) $y$
(b) $y/2$
(c) $2y$
(d) $y/6$
I expanded the first equation to reach $2\sin x(2+\cos x-2\sin x)= \sin y$, but I doubt it leads me anywhere. A little hint would be appreciated. Thanks!
 A: Note that $$\sin(x)+\sin(2x)+\sin(3x)=\sin(2x)\,\big(1+2\cos(x)\big)$$
and
$$\cos(x)+\cos(2x)+\cos(3x)=\cos(2x)\,\big(1+2\cos(x)\big)\,.$$
Thus, for the required equalities to be true, we need
$$1+2\cos(x)=s\in\{-1,+1\}\text{ and }2x=\left\{\begin{array}{ll}
2k\pi+y\,,&\text{if }s=+1\\
(2k+1)\pi-y\,,&\text{if }s=-1\,.
\end{array}\right.$$
for some integer $k$.  This mean
$$x=\left\{\begin{array}{ll}
k\pi+\frac{y}{2}\,,&\text{if }s=+1\\
k\pi+\frac{\pi-y}{2}\,,&\text{if }s=-1\,.
\end{array}\right.$$
The problem statement seems to suggest that (b) is the correct answer (with option $s=+1$ and $k=0$).  However, there is a caveat, as Saucy O'Path mentioned.
To be precise, the possible values of $(x,y)$ are
$$(x,y)=\left(\frac{\pi}{2}+m\pi,(2n+1)\pi\right)\text{ and }(x,y)=\big((2m+1)\pi,(2n+1)\pi\big)$$
where $m,n\in\mathbb{Z}$.
In addition, $x=\pi$ is a solution, but if (b) is true, then $y=2x=2\pi$ would correspond to the value $x=\pi$, but this is not the case.  When $x=\pi$, the only working values of $y$ are odd multiples of $\pi$.  
Conclusion: This is a poorly designed problem, and should be ignored.  None of the provided choices is a (solely) correct answer.  Each except (c) can be correct, e.g., (a) with $(x,y)=(\pi,\pi)$, (b) with $(x,y)=\left(\frac{\pi}{2},\pi\right)$, and (d) with $(x,y)=\left(\frac{\pi}{2},3\pi\right)$. 
A: Hint: $$\sin(x)+\sin82x)+\sin(3x)=\sin(2x)(2\cos(x)+1)=\sin(y)$$
$$\cos(x)+\cos(2x)+\cos(3x)=\cos(2x)(2\cos(x)+1)=\cos(y)$$
from here you will get
$$\tan(2x)=\tan(y)$$ 
con you finish?
A: Draw 3 unit vectors $e_1$, $e_2$ and $e_3$. The angle between these each of these vectors and $x$ axis is $x$, $2x$ and $3x$ correspondingly.
$x$ coordinates of this vectors would be $cos(x)$, $cos(2x)$ and $cos(3x)$.
To find the sum of these coordinates you can first add up the vectors and than the x-coordinate of the sum of these 3 vectors.
So, $cos(x) + cos(2x) + cos(3x)$ and $sin(x) + sin(2x) + sin(3x)$ are the coordinates of $e_1 + e_2 + e_3$ vector!
And it must be coordinates of unit vector $e_y$!
Now it's a geometry problem which looks much easier for me. $e_1 + e_2 + e_3$ should be pointing to the same direction as $e_2$.
$x=y=\pi$ seems to be a solution.
$x=\pi / 2, y=\pi$ is also a solution.
So (a) may hold and (b) also may hold.
A: If $z = e^{ix} = \cos x + i \sin x$
Then we have $z + z^2 + z^3 = e^{iy} = w$ (say)
Divide by $z^2$ to see that
$$z + \frac{1}{z} + 1 = \frac{w}{z^2}$$
The left side is real and thus
$$w = az^2$$
Since $|w| = |z| = 1$ we must have that $|a| = 1$
Thus $$w = \pm z^2$$
This gives rise to two equations:
$$z + z^3 = 0$$
and
$$z + 2z^2 + z^3 = 0$$
I will leave the rest to you. 
And as others said, be careful that $e^{ix}$ is periodic and for the question to make sense you might need to put bounds on $x,y$.
A: Generalization :
Using  How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?,
$$f(n)=\dfrac{\sum_{k=0}^{n-1}\sin (a+k \cdot d) }{\sum_{k=0}^{n-1}\cos (a+k \cdot d)}= \tan \frac{2  a + (n-1)d}2$$
If $a=d=x,$  $$f(n)=\tan\dfrac{x(n+1)}2$$
Here $n=3$
