For which pairs of numbers $x, y$ does the following equality hold:

$$
3 \sin x-4 \cos x=4 y^{2}+4 y+6
$$
I. solution. The right-hand side of the equality can be written in the form $(2 y+1)^{2}+5$, so it is never less than 5 and is only equal to 5 when $y=-1 / 2$. Therefore, the equation has a solution for those $x$ for which

$$
\begin{gathered}
3 \sin x-4 \cos x \geqq 5, \quad \text { that is } \\
3 \sin x \geqq 5+4 \cos x
\end{gathered}
$$

The right-hand side is certainly positive in (2) (since $4 \cos x \geqq-4$), so $\sin x>0$. But then both sides of the inequality are positive, so we can square them. Substituting the identity $\sin ^{2} x=1-\cos ^{2} x$ into the resulting inequality and rearranging, we get the inequality

$$
0 \geqq 16+40 \cos x+25 \cos ^{2} x=(5 \cos x+4)^{2}
$$

From this, it is clear that (3) and thus the equivalent (2) only hold if $\cos x=-\frac{4}{5}$. Therefore, the equation (1) has a solution for those $x$ for which $\sin x>0$ and $\cos x=-\frac{4}{5}$, that is, $x=-\arccos \frac{4}{5}+(2 k+1) \pi \approx 2.498+2 k \pi$ (or in degrees, $x \approx 143.13^{\circ}+k \cdot 360^{\circ}$), where $k$ is an integer. In this case, the value of the left-hand side of (1) is 5, so the right-hand side is also 5, and thus $y=-\frac{1}{2}$. The solutions of the equation are given by the pairs $x=-\operatorname{arc} \cos \frac{4}{5}+(2 k+1) \pi, y=-\frac{1}{2}$ (where $k$ is an integer).

In the course of the solution, we also proved that

$$
3 \sin x-4 \cos x \leqq 5
$$

for all $x$, and equality holds only when $\cos x=-\frac{4}{5}$.

II. solution. The statement mentioned at the end of the first solution can also be proven with a frequently applicable trick. Write (4) in the form

$$
\frac{3}{5} \sin x-\frac{4}{5} \cos x \leqq 1
$$

Since $\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}=1$, there is an angle $\varphi$ whose cosine is $\frac{3}{5}$ and sine is $\frac{4}{5}$ (approximately $\varphi \approx 53.13^{\circ}$). The left-hand side of the inequality can thus be written as $\sin x \cos \varphi-\cos x \sin \varphi=\sin (x-\varphi)$, and this is indeed never greater than 1.

This proves that the left-hand side of our equation is at most 5, and we have seen that the right-hand side is at least 5. Equality can only hold if

$$
\begin{gathered}
4 y^{2}+4 y+6=5 \quad \text { that is } \quad y=-\frac{1}{2} \\
3 \sin x-4 \cos x=5 \quad \text { that is } \quad \sin (x-\varphi)=1
\end{gathered}
$$

so $x=\varphi+\pi / 2+2 k \pi$ (where $k$ is an integer), and $\varphi=\operatorname{arc} \sin 4 / 5 \approx 0.9273$.

Remark. The idea of the second solution generally gives that

$$
-\sqrt{a^{2}+b^{2}} \leqq a \cos x+b \sin x \leqq \sqrt{a^{2}+b^{2}}
$$

for all $x$ (where $a, b$ are fixed real numbers). If $a$ and $b$ are both zero, this is clear. If $a^{2}+b^{2}>0$, we "divide through" by $\sqrt{a^{2}+b^{2}}$, then find (one of the) $\varphi^{\prime}$ such that $\frac{a}{\sqrt{a^{2}+b^{2}}}=\sin \varphi^{\prime}, \frac{b}{\sqrt{a^{2}+b^{2}}}=\cos \varphi^{\prime}$ (such a $\varphi^{\prime}$ exists because $\left(\frac{a}{\sqrt{a^{2}+b^{2}}}\right)^{2}+$ $\left.\left(\frac{b}{\sqrt{a^{2}+b^{2}}}\right)^{2}=1\right)$. Then the inequality takes the form

$$
-1 \leqq \sin \varphi^{\prime} \cos x+\cos \varphi^{\prime} \sin x=\sin \left(\varphi^{\prime}+x\right) \leqq 1
$$

which is obviously true.