Maths Question 4 – JEE-MAIN 2026

Use the trigonometric identities $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$ and $\cos 2\theta =2{\cos }^2\theta -1$. Substitute these into the equation from Step 1:

Factor out $\cos \theta$:

This implies either $\cos \theta =0$ or $4{\cos }^2\theta +4\cos \theta -3=0$.

Case 1: $\cos \theta =0$. For $\theta \in [0,2\pi ]$, the solutions are $\theta =\frac{\pi }{2}$ and $\theta =\frac{3\pi }{2}$.

Case 2: $4{\cos }^2\theta +4\cos \theta -3=0$. Let $c=\cos \theta$. The quadratic equation is $4c^2+4c-3=0$. Using the quadratic formula, $c=\frac{-4\pm \sqrt{4^2-4(4)(-3)}}{2(4)}=\frac{-4\pm \sqrt{16+48}}{8}=\frac{-4\pm \sqrt{64}}{8}=\frac{-4\pm 8}{8}$. This gives two values: $c=\frac{4}{8}=\frac{1}{2}$ or $c=\frac{-12}{8}=-\frac{3}{2}$. Since $\cos \theta$ must be in $[-1,1]$, $c=-\frac{3}{2}$ is not a valid solution. Thus, we only consider $\cos \theta =\frac{1}{2}$. For $\theta \in [0,2\pi ]$, the solutions are $\theta =\frac{\pi }{3}$ and $\theta =\frac{5\pi }{3}$.

The set of all possible values for $\theta$ in $[0,2\pi ]$ is $\{\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{5\pi }{3}\}$. The sum of these values is: