Solution of Triangles & Trig EquationsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Solution of Triangles & Trig Equations: Equals (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
If S={θ∈[−π,π]:cos⁡θcos⁡5θ2=cos⁡7θcos⁡7θ2}S=\left\{\theta\in[-\pi,\pi]:\cos\theta\cos\dfrac{5\theta}{2}=\cos7\theta\cos\dfrac{7\theta}{2}\right\}, then n(S)n(S) equals
Solution
Answer: 19
Step 1: Convert to a single trigonometric equation using product-to-sum Applying cos⁡Acos⁡B=12[cos⁡(A+B)+cos⁡(A−B)]\cos A\cos B = \frac{1}{2}[\cos(A+B)+\cos(A-B)] to both sides:
12 ⁣[cos⁡7θ2+cos⁡3θ2]=12 ⁣[cos⁡21θ2+cos⁡7θ2]\frac{1}{2}\!\left[\cos\frac{7\theta}{2}+\cos\frac{3\theta}{2}\right] = \frac{1}{2}\!\left[\cos\frac{21\theta}{2}+\cos\frac{7\theta}{2}\right]
cos⁡3θ2=cos⁡21θ2\cos\frac{3\theta}{2} = \cos\frac{21\theta}{2}
 Step 2: Factor using sum-to-product
−2sin⁡ ⁣(21θ/2+3θ/22)sin⁡ ⁣(21θ/2−3θ/22)=−2sin⁡6θsin⁡9θ2=0-2\sin\!\left(\frac{21\theta/2+3\theta/2}{2}\right)\sin\!\left(\frac{21\theta/2-3\theta/2}{2}\right) = -2\sin6\theta\sin\frac{9\theta}{2} = 0
 Step 3: Count solutions of sin⁡6θ=0\sin6\theta=0 in [−π,π][-\pi,\pi] θ=nπ6\theta = \dfrac{n\pi}{6} for n=−6,−5,…,6n=-6,-5,\ldots,6: gives 13\mathbf{13} solutions. Step 4: Count solutions of sin⁡9θ2=0\sin\dfrac{9\theta}{2}=0 in [−π,π][-\pi,\pi] θ=2mπ9\theta = \dfrac{2m\pi}{9} for m=−4,−3,…,4m=-4,-3,\ldots,4: gives 9\mathbf{9} solutions. Step 5: Remove overlap using inclusion-exclusion Common elements satisfy both conditions: θ=0\theta=0 and θ=±2π3\theta=\pm\dfrac{2\pi}{3} (verified: 6⋅2π3=4π6\cdot\frac{2\pi}{3}=4\pi, sin⁡4π=0\sin4\pi=0). Total 3\mathbf{3} common values.
n(S)=13+9−3=19n(S) = 13+9-3 = 19
 Answer: 19
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