Question

Find all exact solutions to cos4θ=cos2θ on [0,π).

Answer 1

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:

\[
\boxed{0}.
\]