To fill in the missing values in the table for the function \( f(x) = \cos(6x) \), we'll calculate \( f(x) \) for each given \( x \).
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For \( x = 1 \): \[ f(1) = \cos(6 \cdot 1) = \cos(6) \approx 0.9602 \]
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For \( x = \frac{12}{\pi} \): \[ f\left( \frac{12}{\pi} \right) = \cos\left( 6 \cdot \frac{12}{\pi} \right) = \cos\left( \frac{72}{\pi} \right) \approx \cos(22.9183) \approx -0.9987 \]
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For \( x = \frac{6}{\pi} \): \[ f\left( \frac{6}{\pi} \right) = \cos\left( 6 \cdot \frac{6}{\pi} \right) = \cos\left( \frac{36}{\pi} \right) \approx \cos(11.4592) \approx -0.3220 \]
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For \( x = 0 \): \[ f(0) = \cos(6 \cdot 0) = \cos(0) = 1 \]
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For \( x = \frac{3}{\pi} \): \[ f\left( \frac{3}{\pi} \right) = \cos\left( 6 \cdot \frac{3}{\pi} \right) = \cos\left( \frac{18}{\pi} \right) \approx \cos(5.7296) \approx 0.3556 \]
Now we can summarize the calculated values in the table:
\[ \begin{tabular}{|c|c|} \hline x & f(x) = \cos(6x) \ \hline 1 & \approx 0.9602 \ \frac{12}{\pi} & \approx -0.9987 \ \frac{6}{\pi} & \approx -0.3220 \ 0 & 1 \ \frac{3}{\pi} & \approx 0.3556 \ \hline \end{tabular} \]
Therefore, the filled table looks like this:
\[ \begin{tabular}{|c|c|} \hline x & f(x)=\cos(6x) \ \hline 1 & 0.9602 \ \frac{12}{\pi} & -0.9987 \ \frac{6}{\pi} & -0.3220 \ 0 & 1 \ \frac{3}{\pi} & 0.3556 \ \hline \end{tabular} \]