Solve the following system of equations:

$$
\begin{aligned}
\sin x+\sin (x+y)+\sin (x+2 y) & =\sin \varphi \ldots \\
\cos x+\cos (x+y)+\cos (x+2 y) & =\cos \varphi \ldots
\end{aligned}
$$

Here $\varphi$ represents a given angle.
$$
\begin{aligned}
& \sin x+\sin (x+2 y)=2 \sin (x+y) \cos y \ldots \\
& \cos x+\cos (x+2 y)=2 \cos (x+y) \cos y \ldots
\end{aligned}
$$

Substituting (3) into (1) and (4) into (2), we get

$$
\begin{aligned}
& \sin (x+y)[1+2 \cos y]=\sin \varphi \ldots \\
& \cos (x+y)[1+2 \cos y]=\cos \varphi \ldots
\end{aligned}
$$

Squaring both sides of the last equation and adding them:

$$
\left[\sin ^{2}(x+y)+\cos ^{2}(x+y)\right][1+2 \cos y]^{2}=\sin ^{2} \varphi+\cos ^{2} \varphi
$$

Thus,

$$
(1+2 \cos y)^{2}=1 \quad \text { or } \quad 4 \cos ^{2} y+4 \cos y=0 \ldots
$$

From this:

$$
\cos y=0 \quad \text { and } \quad \cos y=-1
$$

I. If $\cos y=0$, then from (1a) and (2a):

$$
\sin (x+y)=\sin \varphi \quad \text { and } \quad \cos (x+y)=\cos \varphi
$$

Thus, $x+y=\varphi$ and since $\cos y=0$, $y=\frac{\pi}{2}$ or $\frac{3 \pi}{2}$.

Accordingly,

$$
x=\varphi-\frac{\pi}{2} \quad \text { or } \quad x=\varphi-\frac{3 \pi}{2}
$$

II. If $\cos y=-1$, then from (1a) and (2a):

$$
-\sin (x+y)=\sin \varphi \quad \text { and } \quad-\cos (x+y)=\cos \varphi
$$

This is only possible if $x+y=\varphi+\pi$.

Since $\cos y=-1$, $y=\pi$, thus $x=4 \sqrt{2}$.

Miklós Nádler (Kossuth Lajos g. VII. o. Pestszenterzsébet.)[^0]


[^0]:    ${ }^{1}$ Generally: $x+y=\varphi+2 k \pi, y=\frac{\pi}{2}+l \pi$ and $x=\varphi-\frac{\pi}{2}+n \pi$, where $l$ and $n$ are any integers.

    ${ }^{2}$ Generally $x+y=\varphi+(2 k+1) \pi, y=(2 l+1) \pi$ and thus $x=\varphi+2 n \pi$.