\section*{Problem 2 - 031242}

Determine all real values of \(x\) that satisfy the following equation:

\[
\frac{\sin 3 x \cdot \cos \left(\frac{\pi}{3}-4 x\right)+1}{\sin \left(\frac{\pi}{3}-7 x\right)-\cos \left(\frac{\pi}{6}+x\right)+m}=0
\]

Here, \(m\) is a given real number.
}

A fraction is exactly zero if and only if its numerator is zero and its denominator is not zero.

\[
\sin 3 x \cdot \cos \left(\frac{\pi}{3}-4 x\right)+1=0 \quad ; \quad \sin 3 x \cdot \cos \left(\frac{\pi}{3}-4 x\right)=-1
\]

Since the absolute values of sine and cosine are always less than or equal to 1, for a solution, both must take values with an absolute value of 1:

\[
\sin 3 x=1 ; \quad \cos \left(\frac{\pi}{3}-4 x\right)=-1 \quad \text { or } \quad \sin 3 x=-1 ; \quad \cos \left(\frac{\pi}{3}-4 x\right)=1
\]

When do sine and cosine take values with an absolute value of 1?

\[
\begin{aligned}
& \sin y=1 \Leftrightarrow y \in\left\{2 \pi k+\frac{\pi}{2}: k \in \mathbb{N}\right\} \\
& \sin y=-1 \Leftrightarrow y \in\left\{2 \pi k-\frac{\pi}{2}: k \in \mathbb{N}\right\} \\
& \cos z=1 \Leftrightarrow z \in\{2 \pi j: j \in \mathbb{N}\} \\
& \cos z=-1 \Leftrightarrow z \in\{2 \pi j+\pi: j \in \mathbb{N}\}
\end{aligned}
\]

This means for \(y=3 x\) and \(z=\frac{\pi}{3}-4 x\)

\[
\begin{aligned}
\sin 3 x=1 & \Leftrightarrow 3 x \in\left\{\frac{\pi}{2}(4 k+1): k \in \mathbb{N}\right\} \\
& \Leftrightarrow x \in\left\{\frac{\pi}{6}(4 k+1): k \in \mathbb{N}\right\} \\
\sin 3 x=-1 & \Leftrightarrow x \in\left\{\frac{\pi}{6}(4 k-1): k \in \mathbb{N}\right\} \\
\cos \left(\frac{\pi}{3}-4 x\right)=1 & \Leftrightarrow \frac{\pi}{3}-4 x \in\{2 \pi j: j \in \mathbb{N}\} \\
& \Leftrightarrow 4 x \in\left\{\frac{\pi}{3}-2 \pi j: j \in \mathbb{N}\right\} \\
& \Leftrightarrow x \in\left\{\frac{\pi}{12}(1-6 j): j \in \mathbb{N}\right\} \\
\cos \left(\frac{\pi}{3}-4 x\right)=-1 & \Leftrightarrow x \in\left\{\frac{\pi}{12}(-2-6 j): j \in \mathbb{N}\right\}
\end{aligned}
\]

We are looking for pairs \((k, j)\) of integers with the following conditions:

1. Case: \(\sin y=-1 \wedge \cos z=1\) :

\[
\frac{\pi}{6}(4 k-1)=\frac{\pi}{12}(1-6 j) \Rightarrow 2(4 k+3 j)=3
\]

This obviously does not lead to a solution, as the left side of the equation is always divisible by 2, while the right side is never.

2. Case: \(\sin y=1 \wedge \cos z=-1\) :

\[
\frac{\pi}{6}(4 k+1)=\frac{\pi}{12}(-2-6 j) \Rightarrow 4 k+3 j=-2 \Rightarrow
\]

This equation is satisfied for \((k, j)=(1,-2)\) and otherwise only if one adds a multiple of \(\operatorname{kgV}(4,3)=12\) to the term \(4 k\) and subtracts the same value from \(3 j\).

\[
(k, j) \in\{(1+3 l,-2-4 l): l \in \mathbb{Z}\}
\]

From the values of \(k\) found above, the following values for \(x\) result:

\[
x \in\left\{\frac{\pi}{6}(4 k+1): k \in 1+3 l: l \in \mathbb{Z}\right\}=\left\{\frac{\pi}{6}(12 l+5): l \in \mathbb{Z}\right\}
\]

Finally, it must be ensured that the denominator of the fraction in the problem statement is not zero when the numerator is zero.

\[
\begin{gathered}
\sin \left(\frac{\pi}{3}-7 x\right)-\cos \left(\frac{\pi}{6}+x\right)+m= \\
=\sin \left(\frac{\pi}{3}-7 \frac{\pi}{6}(12 l+5)\right)-\cos \left(\frac{\pi}{6}+\frac{\pi}{6}(12 l+5)\right)+m= \\
=\sin -\frac{11 \pi}{2}-\cos \pi+m=1-(-1)+=2+m
\end{gathered}
\]

This means that the denominator has the value \(2+m\) whenever the numerator is zero. If \(m=-2\), the fraction will never be zero; if, on the other hand, \(m \neq-2\), the fraction will be zero exactly when there is an integer \(l\) with \(x=\frac{\pi}{6}(12 l+5)\).