Solve the following equation and system of equations:

(1) $3 \cos x + 4 \sin x = -1.4$

$\{$ (2a) $13 \cos x - 41 \cos y = -45$

$\{(2 b) \quad 13 \sin x + 41 \sin y = 3$.
I. Solution. a) From (1), by rearrangement, squaring, and then substituting $\cos ^{2} x=1-\sin ^{2} x$, we obtain a quadratic equation in $\sin x$:

$$
\begin{gathered}
3 \cos x=-1.4-4 \sin x \\
9 \cos ^{2} x=9-9 \sin ^{2} x=1.96+11.2 \sin x+16 \sin ^{2} x \\
25 \sin ^{2} x+11.2 \sin x-7.04=0
\end{gathered}
$$

Thus, only angles that satisfy

$$
(\sin x)_{1}=-4 / 5=-0.8, \quad(\sin x)_{2}=44 / 125=0.352
$$

can be solutions. Each of these values corresponds to two angles between $0^{\circ}$ and $360^{\circ}$, specifically in the III. and IV., and I. and II. quadrants, respectively, so each value has two cosine values of equal magnitude but opposite signs. Therefore, out of the two pairs of angles, only one in each pair satisfies (1), which we would need to determine by substitution. This can be avoided if we first calculate $\cos x$ from (4) based on (1):

$$
(\cos x)_{1}=+3 / 5=0.6, \quad(\cos x)_{2}=-117 / 125=-0.936
$$

because this way, the determination of $x_{1}$ and $x_{2}$ from the two functions is unique. Based on the sign pairs, $x_{1}$ is in the IV. quadrant, and $x_{2}$ is in the II. quadrant.

$$
x_{1} \approx 306.87^{\circ}+k \cdot 360^{\circ}, \quad x_{2} \approx 159.39^{\circ}+k \cdot 360^{\circ}
$$

where $k$ is an integer. After identifying the quadrants, the determination can be done from either $\sin x$ or $\cos x$, because from (3) and (1) we get that the sum of the squares of the corresponding $\sin x, \cos x$ value pairs is 1.

b) Substituting the expressions for $\cos y$ and $\sin y$ from (2) into the identity $\cos ^{2} y+\sin ^{2} y=1$, we obtain a single-variable equation in $x$. This will be of the same type as (1), because in (2a) and (2b) the ratio of the absolute values of the coefficients is the same: $13 / 41$, so the coefficients of $\cos ^{2} x$ and $\sin ^{2} x$ are equal, and their sum is independent of $x$:

$$
\begin{gathered}
\left(\frac{13 \cos x+45}{41}\right)^{2}+\left(\frac{3-13 \sin x}{41}\right)^{2}=1 \\
169\left(\cos ^{2} x+\sin ^{2} x\right)+1170 \cos x-78 \sin x+2034=1681 \\
195 \cos x-13 \sin x=-87
\end{gathered}
$$

From here, using the method seen in part a),

$$
\begin{aligned}
\sin x_{1} & =12 / 13 \approx 0.9231 \\
\cos x_{1} & =-5 / 13 \approx-0.3846
\end{aligned}
$$

$$
\begin{aligned}
& \sin x_{2}=-1269 / 1469 \approx-0.8639 \\
& \cos x_{2}=-740 / 1469 \approx-0.5037
\end{aligned}
$$

and further from (2a) and (2b),

$$
\begin{array}{ll}
\sin y_{1}=-9 / 41 \approx-0.2195, & \sin y_{2}=1608 / 4633 \approx 0.3471 \\
\cos y_{1}=40 / 41 \approx 0.9756, & \cos y_{2}=4345 / 4633 \approx 0.9378
\end{array}
$$

Thus, the system of equations has two solutions:

$$
\left\{\begin{array} { l } 
{ x _ { 1 } \approx 1 1 2 . 6 4 ^ { \circ } + k \cdot 3 6 0 ^ { \circ } , } \\
{ y _ { 1 } \approx 3 4 7 . 3 2 ^ { \circ } + k \cdot 3 6 0 ^ { \circ } ; }
\end{array} \quad \left\{\begin{array}{l}
x_{2} \approx 239.75^{\circ}+k \cdot 360^{\circ} \\
y_{2} \approx 20.31^{\circ}+k \cdot 360^{\circ}
\end{array}\right.\right.
$$

Kálmán Horváth (Kaposvár, Táncsics M. g. III. o. t.)

II. Solution. a) Thinking of the

$$
\sin \alpha \cos x+\cos \alpha \sin x=\sin (x+\alpha)
$$

identity, we seek a number $\lambda$ such that multiplying (1) by it results in a $\alpha$ angle where the $3 \lambda, 4 \lambda$ coefficients are equal to $\sin \alpha$ and $\cos \alpha$,