
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Rewrite the equation in the form:

$$
\cos 3a \cdot \sin x + (\sin 3a - \sin 7a) \cos x = 0
$$

Consider the following cases:

Case 1: \(\cos 3a \neq 0\) and \(\sin 3a - \sin 7a \neq 0\):
The equation becomes:

$$
\operatorname{tg} x = \frac{\sin 7a - \sin 3a}{\cos 3a}
$$

Solutions are given by:

$$
x_{n} = \operatorname{arctg}\left(\frac{\sin 7a - \sin 3a}{\cos 3a}\right) + \pi n, \, n \in \mathbb{Z}
$$

Case 2: \(\cos 3a \neq 0\) and \(\sin 3a - \sin 7a = 0\):
The equation becomes:

$$
\cos 3a \cdot \sin x = 0
$$

Solutions are given by:

$$
x_{n} = \pi n, \, n \in \mathbb{Z}
$$

Case 3: \(\cos 3a = 0\) and \(\sin 3a - \sin 7a \neq 0\):
The equation becomes:

$$
(\sin 3a - \sin 7a) \cos x = 0
$$

Solutions are given by:

$$
x_{n} = \frac{\pi}{2} + \pi n, \, n \in \mathbb{Z}
$$

Case 4: \(\cos 3a = 0\) and \(\sin 3a - \sin 7a = 0\):
The equation becomes:

$$
0 = 0
$$

In this case, any \( x \) is a solution to the equation.

Determine the values of \( a \) for which:

$$
\left\{
\begin{array}{l}
\cos 3a = 0 \\
\sin 3a - \sin 7a = 0
\end{array}
\right.
$$

Solve \(\cos 3a = 0\):

$$
a = \frac{\pi(2m+1)}{6}, \, m \in \mathbb{Z}
$$

Rewriting \(\sin 3a - \sin 7a = 0\) gives:

$$
-2 \sin 2a \cdot \cos 5a = 0
$$

Solutions are:

$$
a = \frac{\pi n}{2}, \, n \in \mathbb{Z}
$$

and

$$
a = \frac{\pi(2k+1)}{10}, \, k \in \mathbb{Z}
$$

The system becomes:

$$
\left\{
\begin{array}{l}
a = \frac{\pi(2m+1)}{6}, \, m \in \mathbb{Z} \\
a = \frac{\pi n}{2}, \, n \in \mathbb{Z}
\end{array}
\right.
$$

Solve \(3n - 2m = 1\):

$$
\left\{
\begin{array}{l}
n = 2t + 1 \\
m = 3t + 1
\end{array}
\right.
$$

Therefore:

$$
a = \frac{\pi(2t+1)}{2}, \, t \in \mathbb{Z}
$$

Solve \(5m - 3k = -1\):

$$
\left\{
\begin{array}{l}
m = 3t + 1 \\
k = 5t + 2
\end{array}
\right.
$$

Therefore:

$$
a = \frac{\pi(2t+1)}{2}, \, t \in \mathbb{Z}
$$

Both systems yield the same solution:

$$
a = \frac{\pi (2t + 1)}{2}, \, t \in \mathbb{Z}
$$

Present the answer in LaTex format: \boxed{Your answer}
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