Solution to trigonometric system of equations? I am given the following system of trigonometric equations for parameters $x,y \in \mathbb{R}$, with $p:= \sqrt3/2$ and $q:=1/2$:
$$\begin{align}
\cos y 
- \sin\left(\frac\pi6 + px - qy\right) 
- \sin\left(\frac\pi6 + px + qy\right) &= 0 \\[6pt]
\sin y 
+ \cos\left(\frac\pi6 + px - qy\right) 
- \cos\left(\frac\pi6 + px + qy\right) &=0
\end{align}$$
According to WolframAlpha, one can even find some explicit solutions, but I don't quite see how to derive these/all solutions from scratch?
 A: @Claude's answer took a turn I hadn't considered. My comment intended to setup this argument:
$$
\cos y = 2\cos qy\sin\left(\frac\pi6+px\right) \qquad 
\sin y = 2\sin qy\sin\left(\frac\pi6+px\right) \tag1$$
so that
$$1=\cos^2y+\sin^2y=4\left(\cos^2qy+\sin^2qy\right)\sin^2\left(\frac\pi6+px\right)=4\sin^2\left(\frac\pi6+px\right) \tag{2}$$
Thus,
$$\sin\left(\frac\pi6+px\right) = \pm\frac12 =\sin\left(\pm\frac\pi6\right) \tag3$$
and substituting back into $(1)$ gives
$$\cos y = \pm \cos qy \qquad \sin y = \pm \sin qy \tag4$$
where, it should be noted, all "$\pm$"s in $(3)$ and $(4)$ match. From there, solving for $x$ and $y$ is straightforward. $\square$
A: As @Blue commented, rewrite
$$\cos(y)= \sin\left(\frac\pi6 + px - qy\right) + \sin\left(\frac\pi6 + px + qy\right)=\cos (q y) \left(\sqrt{3} \sin (p x)+\cos (p x)\right)$$
$$\sin(y)=\cos\left(\frac\pi6 + px + qy\right)-\cos\left(\frac\pi6 + px - qy\right)=\sin (q y) \left(\cos (p x)\right)-\sqrt{3} \sin (p x))$$
Square both sides, add and simplify to get
$$1=\sqrt{3} \sin (2 p x)-\cos (2 p x)+2$$ which gives a solutions (I omit the modulo's)
$$\left\{\{px\to 0\},\left\{px\to -\pi\right\},\left\{px\to -\frac{\pi }{3
   }\right\},\left\{px\to \frac{2 \pi }{3 }\right\},\left\{px\to {\pi
   }\right\}\right\}$$ For each of these solutions, compute $y$.
