Question

Let \[ \left| \cos \theta \cos (60^\circ - \theta) \cos (60^\circ - \theta) \right| \leq \frac{1}{8}, \quad \theta \in [0, 2\pi] \] Then, the sum of all \( \theta \in [0, 2\pi] \), where \( \cos 3\theta \) attains its maximum value, is:

• A. \( 9\pi \)
• B. \( 18\pi \)
• C. \( 6\pi \)
• D. \( 15\pi \)

Answer 1

The Correct Option is C

Step 1: Simplify the inequality Using the trigonometric identity:

\[ \cos \theta \cos (60^\circ - \theta) \cos (60^\circ + \theta) = \frac{1}{4} \cos 3\theta, \] the inequality reduces to: \[ \left| \frac{1}{4} \cos 3\theta \right| \leq \frac{1}{8}. \]

Simplify further: \[ |\cos 3\theta| \leq \frac{1}{2}. \]

Step 2: Range of \(\cos 3\theta\) The inequality becomes: \[ -\frac{1}{2} \leq \cos 3\theta \leq \frac{1}{2}. \] The maximum value of \(\cos 3\theta\) within this range is \(\frac{1}{2}\). At this value: \[ \cos 3\theta = \frac{1}{2}. \]

Step 3: Solve for \(3\theta\) The general solution for \(\cos 3\theta = \frac{1}{2}\) is: \[ 3\theta = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}. \] Divide through by 3 to solve for \(\theta\): \[ \theta = \frac{2n\pi}{3} \pm \frac{\pi}{9}. \]

Step 4: Possible values of \(\theta\) in \([0, 2\pi]\) For \(\theta \in [0, 2\pi]\), substitute \(n = 0, 1, 2, \dots\) until all possible values of \(\theta\) are found.

• For \(n = 0\):
• For \(n = 1\):
• For \(n = 2\):
• For \(n = 3\):

Thus, the possible values of \(\theta\) are: \[ \theta = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{11\pi}{9}, \frac{13\pi}{9}, \frac{17\pi}{9}. \]

Step 5: Sum of all \(\theta\) The sum of these values is: \[ \text{Sum} = \frac{\pi}{9} + \frac{5\pi}{9} + \frac{7\pi}{9} + \frac{11\pi}{9} + \frac{13\pi}{9} + \frac{17\pi}{9}. \] \[ \text{Sum} = \frac{\pi (1 + 5 + 7 + 11 + 13 + 17)}{9} = \frac{\pi \cdot 54}{9} = 6\pi. \]

Final Answer: Option (3).