JEE Main 2026 — Mathematics Trigonometry

Given the equation: $\sin x (\sin x + \cos x) = a$sinx(sinx+cosx)=a

$$\Rightarrow \sin^2 x + \sin x \cos x = a$$⇒sin2x+sinxcosx=a

Using the double angle formulas $\sin^2 x = \dfrac{1 - \cos 2x}{2}$sin2x=21−cos2x​ and $\sin x \cos x = \dfrac{\sin 2x}{2}$sinxcosx=2sin2x​, we get:

$$\dfrac{1 - \cos 2x}{2} + \dfrac{\sin 2x}{2} = a$$21−cos2x​+2sin2x​=a

$$\Rightarrow \sin 2x - \cos 2x = 2a - 1$$⇒sin2x−cos2x=2a−1

We know that the range of $A\sin \theta + B\cos \theta$Asinθ+Bcosθ is $[-\sqrt{A^2+B^2}, \sqrt{A^2+B^2}]$[−A2+B2​,A2+B2​]. Thus, the range of $\sin 2x - \cos 2x$sin2x−cos2x is $[-\sqrt{2}, \sqrt{2}]$[−2​,2​].

$$\Rightarrow -\sqrt{2} \le 2a - 1 \le \sqrt{2}$$⇒−2​≤2a−1≤2​

$$\Rightarrow 1 - 1.414 \le 2a \le 1 + 1.414$$⇒1−1.414≤2a≤1+1.414

$$\Rightarrow -0.414 \le 2a \le 2.414$$⇒−0.414≤2a≤2.414

$$\Rightarrow -0.207 \le a \le 1.207$$⇒−0.207≤a≤1.207

Since $a \in \mathbb{Z}$a∈Z, the possible values for $a$a are $0$0 and $1$1.

Case 1: $a = 0$a=0

$$\sin x (\sin x + \cos x) = 0$$sinx(sinx+cosx)=0

$\Rightarrow \sin x = 0$⇒sinx=0 or $\sin x + \cos x = 0$sinx+cosx=0

For $x \in [-\pi, \pi]$x∈[−π,π], $\sin x = 0 \Rightarrow x \in \{-\pi, 0, \pi\}$sinx=0⇒x∈{−π,0,π} (3 solutions)

$\sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x \in \{-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\}$sinx+cosx=0⇒tanx=−1⇒x∈{−4π​,43π​} (2 solutions)

Total solutions for $a = 0$a=0 is $5$5.

Case 2: $a = 1$a=1

$$\sin^2 x + \sin x \cos x = 1$$sin2x+sinxcosx=1

$$\Rightarrow \sin x \cos x = 1 - \sin^2 x$$⇒sinxcosx=1−sin2x

$$\Rightarrow \sin x \cos x = \cos^2 x$$⇒sinxcosx=cos2x

$$\Rightarrow \cos x (\sin x - \cos x) = 0$$⇒cosx(sinx−cosx)=0

$\Rightarrow \cos x = 0$⇒cosx=0 or $\sin x - \cos x = 0$sinx−cosx=0

For $x \in [-\pi, \pi]$x∈[−π,π], $\cos x = 0 \Rightarrow x \in \{-\dfrac{\pi}{2}, \dfrac{\pi}{2}\}$cosx=0⇒x∈{−2π​,2π​} (2 solutions)

$\sin x - \cos x = 0 \Rightarrow \tan x = 1 \Rightarrow x \in \{-\dfrac{3\pi}{4}, \dfrac{\pi}{4}\}$sinx−cosx=0⇒tanx=1⇒x∈{−43π​,4π​} (2 solutions)

Total solutions for $a = 1$a=1 is $4$4.

The total number of elements in the set $S$S is $n(S) = 5 + 4 = 9$n(S)=5+4=9.

Answer: $9$9