JEE Main 2024 — Mathematics Trigonometry

We know that $(\cos \theta)\left(\cos( 6 0 ^ { \circ } - \theta ) \left(\cos \left(60^{\circ}+\theta\right)=\frac{1}{4} \cos 3 \theta\right.\right.$(cosθ)(cos(60∘−θ)(cos(60∘+θ)=41​cos3θ So equation reduces to $\left|\frac{1}{4} \cos 3 \theta\right| \leq \frac{1}{8}$​41​cos3θ​≤81​ $\begin{aligned}
& \Rightarrow|\cos 3 \theta| \leq \frac{1}{2} \
& \Rightarrow-\frac{1}{2} \leq \cos 3 \theta \leq \frac{1}{2}
\end{aligned}$$\Rightarrow$maximum value of$maximumvalueof\cos 3 \theta=\frac{1}{2}$, here$,here\begin{aligned}
& \Rightarrow 3 \theta=2 \mathrm{n} \pi \pm \frac{\pi}{3} \
& \theta=\frac{2 \mathrm{n} \pi}{3} \pm \frac{\pi}{9}
\end{aligned}$As$As\theta \in[0,2 \pi]$possible values are$possiblevaluesare\theta=\left{\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9}\right}$Whose sum is$Whosesumis\frac{\pi}{9}+\frac{5 \pi}{9}+\frac{7 \pi}{9}+\frac{11 \pi}{9}+\frac{13 \pi}{9}+\frac{17 \pi}{9}=\frac{54 \pi}{9}=6 \pi$