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

Answer: (C)

$6\pi$

Answer 2

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).