JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 23

Products to sums.
$$\cos\theta\cos\tfrac{5\theta}2=\tfrac12\left[\cos\tfrac{7\theta}2+\cos\tfrac{3\theta}2\right],\qquad\cos7\theta\cos\tfrac{7\theta}2=\tfrac12\left[\cos\tfrac{21\theta}2+\cos\tfrac{7\theta}2\right].$$

So $\cos\tfrac{3\theta}2=\cos\tfrac{21\theta}2$, which gives $9\theta=2n\pi$ or $12\theta=2n\pi$.

In $[-\pi,\pi]$.

• $\theta=\tfrac{2n\pi}9$, $n=-4,\dots,4$: 9 values.
• $\theta=\tfrac{n\pi}6$, $n=-6,\dots,6$: 13 values.
• Common to both: $0,\pm\tfrac{2\pi}3$.

Count: $9+13-3=19$.

Answer: 19