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[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

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[ACTION NAME]

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# Your action step 3

...

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Suppose that \( T \) is the period of the function \( y = f(x) \). From the given condition, we have:

\[ 
\cos x = f(x) - 3 f(x - \pi) = f(x - T) - 3 f(x - T - \pi) = \cos (x - T)
\]

Since this equality holds for all \( x \), the number \( T \) is the period of the function \( y = \cos x \), i.e., \( T = 2 \pi n \), where \( n \) is some natural number. From the given equation, we get:

\[ 
\begin{gathered}
f(x)=3f(x-\pi)+\cos x = 3(3 f(x-2\pi) - \cos x) + \cos x = \\
=3(3(3 f(x-3\pi) + \cos x) - \cos x) + \cos x = \\
=\ldots=3(3(\ldots(3 f(x-2\pi n) - \cos x) + \ldots + \cos x) - \cos x) + \cos x = \\
=3^{2n} f(x) - \cos x \left(3^{2n-1} - 3^{2n-2} + 3^{2n-3} - \ldots + 3 - 1\right)
\end{gathered}
\]

Then

\[ 
f(x) \left(1 - 3^{2n}\right) = \cos x \left(1 - 3 + 3^2 - \ldots - 3^{2n-1}\right) = \cos x \cdot \frac{1 - 3^{2n}}{1 + 3}
\]

Therefore, \( f(x) = \frac{1}{4} \cos x \).

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