To solve the equation \(\cos 4\theta = \cos 2\theta\) on the interval \([0, \pi)\), we will first use the identity that says that if \(\cos A = \cos B\), then:
\[ A = 2n\pi + B \quad \text{or} \quad A = -2n\pi + B \quad \text{for some integer } n. \]
Let \(A = 4\theta\) and \(B = 2\theta\). We have two cases to consider:
Case 1: \[ 4\theta = 2n\pi + 2\theta \] Rearranging gives: \[ 4\theta - 2\theta = 2n\pi \] \[ 2\theta = 2n\pi \] Dividing by 2 yields: \[ \theta = n\pi. \] In the interval \([0, \pi)\), we can have:
- For \(n = 0\), \(\theta = 0\).
- For \(n = 1\), \(\theta = \pi\) (not included as \(\pi\) is not in the interval).
Thus, from Case 1, we have one solution: \(\theta = 0\).
Case 2: \[ 4\theta = -2n\pi + 2\theta \] Rearranging this equation gives: \[ 4\theta - 2\theta = -2n\pi \] \[ 2\theta = -2n\pi \] Dividing by 2 yields: \[ \theta = -n\pi. \] For \(n = 0\), this gives \(\theta = 0\) (already counted). For \(n = 1\), \(\theta = -\pi\) (not in the interval). For \(n = 2\), we have \(\theta = -2\pi\) (also not in the interval).
Thus, from Case 2, we do not find any additional solutions.
Next, we can verify the solution we found. The only solution from both cases in the interval \([0, \pi)\) is:
\[ \theta = 0. \]
To confirm, substitute \(\theta = 0\) back into the original equation: \[ \cos 4(0) = \cos 2(0) \implies \cos(0) = \cos(0) \implies 1 = 1. \]
Both sides are equal, confirming that our solution is valid.
Thus, the exact solution to the equation \(\cos 4\theta = \cos 2\theta\) on the interval \([0, \pi)\) is:
\[ \boxed{0}. \]