Solution of Triangles & Trig EquationsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

sin x (sin x + cos x) = a, Integer a: n(S) = 9 | JEE Main 2026

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

Question
Let S={x∈[−π,π]:sin⁡x(sin⁡x+cos⁡x)=a, a∈Z}S=\{x\in[-\pi,\pi]:\sin x(\sin x+\cos x)=a,\ a\in\mathbb{Z}\}. Then n(S)n(S) is equal to
A33
B66
C77
D99correct
Solution
Step 1: Expand and rewrite:
sin⁡x(sin⁡x+cos⁡x)=sin⁡2x+sin⁡xcos⁡x=1−cos⁡2x2+sin⁡2x2=1−cos⁡2x+sin⁡2x2.\sin x(\sin x+\cos x)=\sin^2x+\sin x\cos x=\frac{1-\cos2x}{2}+\frac{\sin2x}{2}=\frac{1-\cos2x+\sin2x}{2}.
 Its range is [1−22, 1+22]≈[−0.207, 1.207]\left[\dfrac{1-\sqrt2}{2},\ \dfrac{1+\sqrt2}{2}\right]\approx[-0.207,\ 1.207], so the integer values possible are a=0a=0 and a=1a=1. Step 2: Case a=0a=0: sin⁡x(sin⁡x+cos⁡x)=0\sin x(\sin x+\cos x)=0. - sin⁡x=0⇒x=−π, 0, π\sin x=0\Rightarrow x=-\pi,\ 0,\ \pi (3 solutions). - sin⁡x+cos⁡x=0⇒tan⁡x=−1⇒x=−π4, 3π4\sin x+\cos x=0\Rightarrow \tan x=-1\Rightarrow x=-\dfrac{\pi}{4},\ \dfrac{3\pi}{4} (2 solutions). Total for a=0a=0: 55 solutions. Step 3: Case a=1a=1: 1−cos⁡2x+sin⁡2x=2⇒sin⁡2x−cos⁡2x=11-\cos2x+\sin2x=2\Rightarrow \sin2x-\cos2x=1. Squaring after rearranging leads to sin⁡4x=0\sin4x=0, whose admissible roots in [−π,π][-\pi,\pi] (satisfying the original) are
x=−3π4, −π4, π4, π2 ⇒ 4 solutions.x=-\frac{3\pi}{4},\ -\frac{\pi}{4},\ \frac{\pi}{4},\ \frac{\pi}{2}\ \Rightarrow\ 4\ \text{solutions}.
 Step 4: Total:
n(S)=5+4=9.n(S)=5+4=9.
 Correct answer: (4)
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