Question

If \( A \) is the solution set of the equation \( \cos^2 x = \cos^2 \frac{\pi}{6} \) and \( B \) is the solution set of the equation \( \cos^2 x = \log_{10} P \) where \( P + \frac{16}{P} = 10 \), then \( B - A = ? \)

• A. \( \left\{ x \in \mathbb{R} \mid x = 2n\pi + \frac{\pi}{4}, 2n\pi \pm \frac{\pi}{3}, n = 0,1,2,3, \dots \right\} \)
• B. \( \left\{ x \in \mathbb{R} \mid x = 2n\pi + \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n = 0,1,2,3, \dots \right\} \)
• C. \( \left\{ x \in \mathbb{R} \mid x = 2n\pi + \frac{\pi}{6}, 2n\pi \pm \frac{\pi}{12}, n = 0,1,2,3, \dots \right\} \)
• D. \( \left\{ x \in \mathbb{R} \mid x = 2n\pi + \frac{\pi}{8}, 2n\pi \pm \frac{\pi}{16}, n = 0,1,2,3, \dots \right\} \)

Hint:

When solving trigonometric equations involving squared functions, always consider both positive and negative values of the cosine function. Also, when given logarithmic expressions, solve for possible values first before substituting into trigonometric identities.

Answer 1

The Correct Option is B

We begin by solving the given trigonometric equations: 

Step 1: Find the solution set \( A \) for \( \cos^2 x = \cos^2 \frac{\pi}{6} \) 
Since: \[ \cos^2 \frac{\pi}{6} = \frac{3}{4} \] The general solution for \( \cos^2 x = k \) is: \[ x = 2n\pi \pm \alpha, \quad \text{where } \alpha = \cos^{-1} \left(\sqrt{\frac{3}{4}}\right) = \frac{\pi}{6} \] Thus, the solution set \( A \) is: \[ A = \left\{ x \mid x = 2n\pi \pm \frac{\pi}{6}, n \in \mathbb{Z} \right\} \] 

Step 2: Find the solution set \( B \) for \( \cos^2 x = \log_{10} P \) 
We are given: \[ P + \frac{16}{P} = 10 \] Multiplying both sides by \( P \): \[ P^2 - 10P + 16 = 0 \] Solving for \( P \): \[ P = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm 6}{2} \] \[ P = \frac{16}{2} = 8, \quad P = \frac{4}{2} = 2 \] Since: \[ \cos^2 x = \log_{10} P \] we substitute: \[ \cos^2 x = \log_{10} 2 \] which corresponds to: \[ x = 2n\pi \pm \frac{\pi}{3}, \quad x = 2n\pi \pm \frac{2\pi}{3} \] Thus, the solution set \( B \) is: \[ B = \left\{ x \mid x = 2n\pi \pm \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z} \right\} \] 

Step 3: Compute \( B - A \) 
Comparing the solution sets, we find: \[ B - A = \left\{ x \mid x = 2n\pi + \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z} \right\} \] Thus, the correct answer is: \[ \mathbf{\left\{ x \in \mathbb{R} \mid x = 2n\pi + \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n = 0,1,2,3, \dots \right\}} \]