Question:

If \(\cos 4x = \cos 3x\), find the general solution for \(x\).

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For \( \cos A = \cos B \), avoid dividing by terms that might be zero.
Always use the general form \( A = 2n\pi \pm B \) to ensure all possible roots are captured in the general solution.
Updated On: Apr 13, 2026
  • \( x = 2n\pi \) or \( x = \frac{2n\pi}{7} \), where \( n \in \mathbb{Z} \)
  • \( x = n\pi \) or \( x = \frac{n\pi}{7} \), where \( n \in \mathbb{Z} \)
  • \( x = \frac{2n\pi}{7} \) only
  • \( x = 2n\pi \) only
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
The question asks for the general solution of a trigonometric equation of the form \( \cos \theta = \cos \alpha \).

Step 2: Key Formula or Approach:

The general solution for \( \cos \theta = \cos \alpha \) is given by \( \theta = 2n\pi \pm \alpha \), where \( n \) is any integer (\( n \in \mathbb{Z} \)).

Step 3: Detailed Explanation:

Given the equation:
\[ \cos 4x = \cos 3x \]
Using the general solution formula, we have:
\[ 4x = 2n\pi \pm 3x \]
This gives us two cases to consider:

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 \( x = \frac{2n\pi}{7} \).
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