Question

If \(\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0\) for \(0 \le \theta \le \pi\), then \(\theta = \):

• A. \(\frac{\pi}{3}, \frac{\pi}{6}, \frac{3\pi}{4}, \frac{5\pi}{6}, \frac{7\pi}{6} \)
• B. \(\frac{\pi}{2}, \frac{\pi}{6}, \frac{3\pi}{4}, \frac{7\pi}{6} \)
• C. \(\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}, \frac{3\pi}{4}, \frac{5\pi}{6} \)
• D. \(\frac{\pi}{2}, \frac{\pi}{4}, \frac{3\pi}{5}, \frac{\pi}{6} \)

Hint:

Use known factorization: \(\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 4\cos 3\theta \cos 2\theta \cos \theta\).

Answer 1

The Correct Option is C

Concept: We use trigonometric factorization of multiple-angle cosine expressions.

Step 1: Use standard factorization identity.
The given expression can be factorized as: \[ \cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 4 \cos 3\theta \cos 2\theta \cos \theta \]

Step 2: Set each factor equal to zero.
\[ 4 \cos 3\theta \cos 2\theta \cos \theta = 0 \] So, \[ \cos 3\theta = 0 \quad \text{or} \quad \cos 2\theta = 0 \quad \text{or} \quad \cos \theta = 0 \]

Step 3: Find values in \( [0,\pi] \).
- \(\cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2}\) - \(\cos 2\theta = 0 \Rightarrow 2\theta = \frac{\pi}{2}, \frac{3\pi}{2}\Rightarrow \theta = \frac{\pi}{4}, \frac{3\pi}{4}\) - \(\cos 3\theta = 0 \Rightarrow 3\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}\Rightarrow \theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\) Final Answer: \[ \boxed{\theta = \frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}, \frac{3\pi}{4}, \frac{5\pi}{6}} \]