Home / doMath / Solution of Triangles & Trig Equations Solution of Triangles & Trig Equations medium PYQ · 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 θ cos 5 θ 2 = cos 7 θ cos 7 θ 2 } S=\left\{\theta\in[-\pi,\pi]:\cos\theta\cos\dfrac{5\theta}{2}=\cos7\theta\cos\dfrac{7\theta}{2}\right\} S = { θ ∈ [ − π , π ] : cos θ cos 2 5 θ = cos 7 θ cos 2 7 θ } , then n ( S ) n(S) n ( S ) equals Solution Answer: 19
Step 1: Convert to a single trigonometric equation using product-to-sum
Applying cos A cos B = 1 2 [ cos ( A + B ) + cos ( A − B ) ] \cos A\cos B = \frac{1}{2}[\cos(A+B)+\cos(A-B)] cos A cos B = 2 1 [ cos ( A + B ) + cos ( A − B )] to both sides:
1 2 [ cos 7 θ 2 + cos 3 θ 2 ] = 1 2 [ cos 21 θ 2 + cos 7 θ 2 ] \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] 2 1 [ cos 2 7 θ + cos 2 3 θ ] = 2 1 [ cos 2 21 θ + cos 2 7 θ ]
cos 3 θ 2 = cos 21 θ 2 \cos\frac{3\theta}{2} = \cos\frac{21\theta}{2} cos 2 3 θ = cos 2 21 θ
Step 2: Factor using sum-to-product
− 2 sin ( 21 θ / 2 + 3 θ / 2 2 ) sin ( 21 θ / 2 − 3 θ / 2 2 ) = − 2 sin 6 θ sin 9 θ 2 = 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 − 2 sin ( 2 21 θ /2 + 3 θ /2 ) sin ( 2 21 θ /2 − 3 θ /2 ) = − 2 sin 6 θ sin 2 9 θ = 0
Step 3: Count solutions of sin 6 θ = 0 \sin6\theta=0 sin 6 θ = 0 in [ − π , π ] [-\pi,\pi] [ − π , π ]
θ = n π 6 \theta = \dfrac{n\pi}{6} θ = 6 nπ for n = − 6 , − 5 , … , 6 n=-6,-5,\ldots,6 n = − 6 , − 5 , … , 6 : gives 13 \mathbf{13} 13 solutions.
Step 4: Count solutions of sin 9 θ 2 = 0 \sin\dfrac{9\theta}{2}=0 sin 2 9 θ = 0 in [ − π , π ] [-\pi,\pi] [ − π , π ]
θ = 2 m π 9 \theta = \dfrac{2m\pi}{9} θ = 9 2 mπ for m = − 4 , − 3 , … , 4 m=-4,-3,\ldots,4 m = − 4 , − 3 , … , 4 : gives 9 \mathbf{9} 9 solutions.
Step 5: Remove overlap using inclusion-exclusion
Common elements satisfy both conditions: θ = 0 \theta=0 θ = 0 and θ = ± 2 π 3 \theta=\pm\dfrac{2\pi}{3} θ = ± 3 2 π (verified: 6 ⋅ 2 π 3 = 4 π 6\cdot\frac{2\pi}{3}=4\pi 6 ⋅ 3 2 π = 4 π , sin 4 π = 0 \sin4\pi=0 sin 4 π = 0 ). Total 3 \mathbf{3} 3 common values.
n ( S ) = 13 + 9 − 3 = 19 n(S) = 13+9-3 = 19 n ( S ) = 13 + 9 − 3 = 19
Answer: 19 Solve more, learn faster Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.