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?
Best Answer
@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. $\square$