3. In the set of real numbers, solve the equation

$$
1+\sin \frac{x+\pi}{5} \cdot \sin \frac{x-\pi}{11}=0 .
$$

The written part of the school round in category A takes place

## on Tuesday, December 6, 2005

so that it starts in the morning and the participants have 4 hours of pure time to solve the problems. For each problem, a participant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies, school MF tables, and calculators without a graphic display. These details will be communicated to the students before the start of the competition.
3. Since all values of the sine function lie in the interval $\langle-1,1\rangle$, the product of two sine values equals -1 only when one value is 1 and the other value is -1. A number $x \in \mathbb{R}$ is thus a solution to the given equation if and only if there exist integers $k, l \in \mathbb{Z}$ such that the following pairs of equations hold:

$$
\left\{\begin{array} { l } 
{ \frac { x + \pi } { 5 } = \frac { \pi } { 2 } + 2 k \pi } \\
{ \frac { x - \pi } { 1 1 } = - \frac { \pi } { 2 } + 2 l \pi }
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
\frac{x+\pi}{5}=-\frac{\pi}{2}+2 k \pi \\
\frac{x-\pi}{11}=\frac{\pi}{2}+2 l \pi
\end{array}\right.\right.
$$

Solving these linear equations, we obtain the expressions

$$
\left\{\begin{array} { l } 
{ x = \frac { 3 \pi } { 2 } + 1 0 k \pi } \\
{ x = - \frac { 9 \pi } { 2 } + 2 2 l \pi }
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
x=-\frac{7 \pi}{2}+10 k \pi \\
x=\frac{13 \pi}{2}+22 l \pi
\end{array}\right.\right.
$$

We will now find all pairs of integers $(k, l)$ for which

$$
\frac{3 \pi}{2}+10 k \pi=-\frac{9 \pi}{2}+22 l \pi, \quad \text { or } \quad-\frac{7 \pi}{2}+10 k \pi=\frac{13 \pi}{2}+22 l \pi .
$$

By simplifying these equations (including dividing by $2 \pi$), we get

$$
5 k+3=11 l, \quad \text { or } \quad 5 k-5=11 l \text {. }
$$

Rewriting the first equation as $5(k-6)=11(l-3)$, and considering the divisibility by the coprime numbers 5 and 11, we find that all integer solutions to this equation are of the form $k=6+11 n$ and $l=3+5 n$, where $n \in \mathbb{Z}$. Substituting into the corresponding formula for $x$, we obtain the first set of solutions

$$
x=\frac{3 \pi}{2}+10 k \pi=\frac{3 \pi}{2}+10(6+11 n) \pi=61.5 \pi+110 n \pi .
$$

Similarly, from the second equation $5 k-5=11 l$ rewritten as $5(k-1)=11 l$, we find that $k=1+11 n, l=5 n$ for any $n \in \mathbb{Z}$, so the second set of solutions is given by

$$
x=-\frac{7 \pi}{2}+10 k \pi=-\frac{7 \pi}{2}+10(1+11 n) \pi=6.5 \pi+110 n \pi
$$

Summary: All solutions to the given equation are given by the formulas

$$
x=61.5 \pi+110 n \pi \quad \text { and } \quad x=6.5 \pi+110 n \pi, \quad \text { where } n \in \mathbb{Z} .
$$

Since $61.5-6.5=55=\frac{110}{2}$, all solutions can be written in a single formula

$$
x=6.5 \pi+55 n \pi, \quad \text { where } n \in \mathbb{Z} .
$$

Alternative solution. Using the trigonometric identity

$$
\sin A \sin B=\frac{\cos (A-B)-\cos (A+B)}{2}
$$

the equation $1+\sin A \sin B=0$ can be rewritten as

$$
\cos (A+B)-\cos (A-B)=2 \text {. }
$$

Considering the range of the cosine function, the last equation is satisfied if and only if $\cos (A+B)=1$ and $\cos (A-B)=-1$. For the fractions $A, B$ from the original equation, we get the system of equations

$$
\begin{aligned}
& A+B=\frac{x+\pi}{5}+\frac{x-\pi}{11}=2 k \pi, \\
& A-B=\frac{x+\pi}{5}-\frac{x-\pi}{11}=\pi+2 l \pi,
\end{aligned}
$$

which must hold for suitable numbers $k, l \in \mathbb{Z}$. Adding and subtracting these equations, we get

$$
\frac{x+\pi}{5}=\frac{\pi}{2}+(k+l) \pi \quad \text { and } \quad \frac{x-\pi}{11}=-\frac{\pi}{2}+(k-l) \pi
$$

from which we can express the unknown $x$ in two ways:

$$
x=\frac{3 \pi}{2}+5(k+l) \pi=-\frac{9 \pi}{2}+11(k-l) \pi .
$$

It is easy to see that the numbers $k, l$ are related by the condition $3(k-1)=8 l$, which means that $l=3 n$ and $k=8 n+1$ for suitable $n \in \mathbb{Z}$. Substituting into the formula for $x$, we arrive at the same expression

$$
x=\frac{13 \pi}{2}+55 n \pi=6.5 \pi+55 n \pi
$$

as in the first solution.

For a complete solution, award 6 points, with 1 point for the consideration of the values of the trigonometric functions sine or cosine and another 2 points for setting up analogous relations for the sought solution $x$. Award the remaining 3 points for expressing $x$ in the form (1) or (2).

[^0]:    1 The square root of a non-negative integer is either an integer or an irrational number.

    2 If $n$ is an integer and $n^{2}$ is divisible by five, then $n$ is also divisible by five.