Question

Find the general solution for \(x\) if \( \cos 4x = \cos 3x \).

• A. \(x = 2n\pi \)
• B. \(x = \frac{2n\pi}{7} \)
• C. \(x = \frac{2n\pi}{7}, \; 2n\pi \)
• D. \(x = \frac{n\pi}{7} \)

Hint:

Whenever you encounter \( \cos A = \cos B \), remember the identity \(A = 2n\pi \pm B\). This automatically generates two sets of solutions which must both be considered.

Answer 1

The Correct Option is C

Concept: A useful trigonometric identity is: \[ \cos A = \cos B \] This implies \[ A = 2n\pi \pm B \qquad (n \in \mathbb{Z}) \] Thus, we solve the equation by considering both possible cases.
Step 1: {Apply the identity \( \cos A = \cos B \).} Given \[ \cos 4x = \cos 3x \] This gives two cases: \[ 4x = 2n\pi + 3x \] or \[ 4x = 2n\pi - 3x \]
Step 2: {Solve the first case.} \[ 4x = 2n\pi + 3x \] \[ x = 2n\pi \]
Step 3: {Solve the second case.} \[ 4x = 2n\pi - 3x \] \[ 7x = 2n\pi \] \[ x = \frac{2n\pi}{7} \]
Step 4: {Combine the solutions.} Thus, the general solution is \[ x = 2n\pi \quad \text{or} \quad x = \frac{2n\pi}{7} \] where \( n \in \mathbb{Z} \).