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

Step 1: Understanding the Question:
The equation is a trigonometric equality of cosine functions.
To find the general solution, we must account for the periodicity and symmetry of the cosine graph across all quadrants.
Step 2: Key Formula or Approach:
The general solution for \( \cos \theta = \cos \alpha \) is:
\[ \theta = 2n\pi \pm \alpha, \text{ where } n \in \mathbb{Z} \]
Step 3: Detailed Explanation:
Given the equation:
\[ \cos 4x = \cos 3x \]
Applying the general formula where \( \theta = 4x \) and \( \alpha = 3x \):
\[ 4x = 2n\pi \pm 3x \]
This gives two separate cases:
Case 1: Positive sign
\[ 4x = 2n\pi + 3x \]
\[ 4x - 3x = 2n\pi \]
\[ x = 2n\pi \]
Case 2: Negative sign
\[ 4x = 2n\pi - 3x \]
\[ 4x + 3x = 2n\pi \]
\[ 7x = 2n\pi \]
\[ x = \frac{2n\pi}{7} \]
Step 4: Final Answer:
Combining both cases, the general solution is \(x = 2n\pi \) or \( x = \frac{2n\pi}{7} \).