
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
Let $(a, b)$ be a pair of numbers that satisfy the conditions of the problem. Then the equality is valid for any $x$, including $x = \pi$ and $x = 2\pi$. Substituting these values of $x$ gives us the system:  
$$
\begin{cases}
-2a + b^2 = \cos(\pi a + b^2) - 1 \\
b^2 = \cos(2\pi a + b^2) - 1
\end{cases}
$$

From the second equality, we deduce that $b^2 \leq 0$, which implies $b = 0$, and thus, $\cos 2\pi a = 1 \Leftrightarrow a = n \in \mathbb{N} \cup \{0\}$. From the first equality, we obtain $\cos \pi a = 1 - 2a \Rightarrow -1 \leq 1 - 2a \leq 1 \Leftrightarrow 0 \leq a = n \leq 1$. Therefore, the only pairs of numbers that satisfy the necessary condition of the problem are $a = 0, b = 0$ and $a = 1, b = 0$. We verify that for these pairs, the equality holds for all $x$. When $a = b = 0$, we have $\cos 0 = 1$, which is true. When $a = 1$ and $b = 0$, we have $\cos x = \cos x$, which is also true.

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