Question

Number of solutions of \( \sqrt{3} \cos 2\theta + 8 \cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi] \) is:

• A. 4
• B. 0
• C. 3
• D. 5

Hint:

When counting solutions in a large range, draw a simple cosine wave graph. For any value $k$ where $|k|<1$, $\cos \theta = k$ has exactly 2 solutions in every $2\pi$ period.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We solve the trigonometric equation by converting it into a quadratic equation in terms of \( \cos \theta \). Then, we check the number of valid roots within the given interval.
Step 2: Key Formula or Approach:
Use the identity: \( \cos 2\theta = 2\cos^2 \theta - 1 \).
Step 3: Detailed Explanation:
Substitute the identity into the original equation: \[ \sqrt{3}(2\cos^2 \theta - 1) + 8 \cos \theta + 3\sqrt{3} = 0 \] \[ 2\sqrt{3}\cos^2 \theta + 8 \cos \theta + 2\sqrt{3} = 0 \] Divide the entire equation by 2: \[ \sqrt{3}\cos^2 \theta + 4 \cos \theta + \sqrt{3} = 0 \] Using the quadratic formula \( \cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ \cos \theta = \frac{-4 \pm \sqrt{16 - 12}}{2\sqrt{3}} = \frac{-4 \pm 2}{2\sqrt{3}} \] The roots are: \( \cos \theta = \frac{-2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}} \) and \( \cos \theta = \frac{-6}{2\sqrt{3}} = -\sqrt{3} \). Since \( |-\sqrt{3}|>1 \), it is rejected. Now, find solutions for \( \cos \theta = -1/\sqrt{3} \) in \( [-3\pi, 2\pi] \): 1. In \( [-3\pi, -\pi] \): 2 solutions. 2. In \( [-\pi, \pi] \): 2 solutions. 3. In \( [\pi, 2\pi] \): 1 solution. Total = 5 solutions.
Step 4: Final Answer:
The total number of solutions is 5.