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

S=\left\{\theta\in[-\pi,\pi]:\cos\theta\cos\dfrac{5\theta}{2}=\cos7\theta\cos\dfrac{7\theta}{2}\right\}

, then

n(S)

equals

Solution

Answer: 19

Step 1: Convert to a single trigonometric equation using product-to-sum Applying

\cos A\cos B = \frac{1}{2}[\cos(A+B)+\cos(A-B)]

to both sides:

\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\frac{3\theta}{2} = \cos\frac{21\theta}{2}

Step 2: Factor using sum-to-product

-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

\sin6\theta=0

[-\pi,\pi]

\theta = \dfrac{n\pi}{6}

for

n=-6,-5,\ldots,6

: gives

\mathbf{13}

solutions. Step 4: Count solutions of

\sin\dfrac{9\theta}{2}=0

[-\pi,\pi]

\theta = \dfrac{2m\pi}{9}

for

m=-4,-3,\ldots,4

: gives

\mathbf{9}

solutions. Step 5: Remove overlap using inclusion-exclusion Common elements satisfy both conditions:

\theta=0

and

\theta=\pm\dfrac{2\pi}{3}

(verified:

6\cdot\frac{2\pi}{3}=4\pi

\sin4\pi=0

). Total

\mathbf{3}

common values.

n(S) = 13+9-3 = 19

Answer: 19

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