Animated Solution for Mathematics - Trigonometry: Let $|\cos \theta \cos (60^{\circ }-\theta )\cos (60^{\circ }+\theta )|\le \frac{1}{8},\theta \in [0,2\pi ]$. Then, the sum of all $\theta \in [0,2\pi ]$, where $\cos 3\theta$ attains its maximum value, is :

Welcome, future engineer. Today, we are not just solving a trigonometric inequality; we are uncovering a hidden symmetry. In the JEE Advanced exam, you will often encounter expressions that look intimidating, designed to make you waste time on brute-force expansion.

Look closely at the expression:

If you try to expand $\cos (60^{\circ }-\theta )$ and $\cos (60^{\circ }+\theta )$ using the compound angle formula, you will eventually get there, but you will be exhausted. Instead, let us invoke the power of the triple angle identity.

This specific product is a classic. It is a beautiful, compact identity:

By recognizing this, we instantly transform a complex product into a single, manageable term. The inequality now reads $|\frac{1}{4}\cos 3\theta |\le \frac{1}{8}$. Multiplying by 4, we arrive at the clean, elegant inequality:

Now, visualize the graph of $\cos 3\theta$. It oscillates between $-1$ and $1$. But our inequality $|\cos 3\theta |\le \frac{1}{2}$ acts like a pair of scissors, cutting off the peaks and valleys.

We are restricted to the region where the function is trapped between $-\frac{1}{2}$ and $\frac{1}{2}$. The question asks for the sum of all $\theta$ where $\cos 3\theta$ attains its maximum value. Within our allowed band, the maximum value is clearly $\frac{1}{2}$.

So, we are solving for:

We know that $\cos (\frac{\pi }{3})=\frac{1}{2}$. Using the general solution for $\cos X=\cos \alpha$, which is $X=2n\pi \pm \alpha$, we set $3\theta =2n\pi \pm \frac{\pi }{3}$.

Dividing by 3, we get our master equation:

Now, we systematically test integer values of $n$ to find all $\theta$ in the interval $[0,2\pi ]$:

For $n=0$: $\theta =\pm \frac{\pi }{9}$. Only $\frac{\pi }{9}$ is in the range. For $n=1$: $\theta =\frac{2\pi }{3}\pm \frac{\pi }{9}$. This gives $\frac{5\pi }{9}$ and $\frac{7\pi }{9}$. For $n=2$: $\theta =\frac{4\pi }{3}\pm \frac{\pi }{9}$. This gives $\frac{11\pi }{9}$ and $\frac{13\pi }{9}$. For $n=3$: $\theta =2\pi \pm \frac{\pi }{9}$. Only $2\pi -\frac{\pi }{9}=\frac{17\pi }{9}$ is in the range.

We have our set of values: $\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9},\frac{11\pi }{9},\frac{13\pi }{9},\frac{17\pi }{9}$. Summing these up, we get:

This is the beauty of mathematics. We started with a complex inequality and, through the lens of identity and symmetry, arrived at a clean, integer-multiple of $\pi$. The final answer is $6\pi$.