JEE Main 2026 — Trigonometric Equations Question with Solution

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

$\Rightarrow {\sin }^{2}x+\sin x\cos x=a$

Using the double angle formulas ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ and $\sin x\cos x=\frac{\sin 2x}{2}$, we get:

$\frac{1-\cos 2x}{2}+\frac{\sin 2x}{2}=a$

$\Rightarrow \sin 2x-\cos 2x=2a-1$

We know that the range of $A\sin \theta +B\cos \theta$ is $[-\sqrt{A^2+B^2},\sqrt{A^2+B^2}]$. Thus, the range of $\sin 2x-\cos 2x$ is $[-\sqrt{2},\sqrt{2}]$.

$\Rightarrow -\sqrt{2}\le 2a-1\le \sqrt{2}$

$\Rightarrow 1-1.414\le 2a\le 1+1.414$

$\Rightarrow -0.414\le 2a\le 2.414$

$\Rightarrow -0.207\le a\le 1.207$

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

Case 1: $a=0$
$\sin x(\sin x+\cos x)=0$
$\Rightarrow \sin x=0$ or $\sin x+\cos x=0$
For $x\in [-\pi ,\pi ]$, $\sin x=0\Rightarrow x\in \{-\pi ,0,\pi \}$ (3 solutions)
$\sin x+\cos x=0\Rightarrow \tan x=-1\Rightarrow x\in \{-\frac{\pi }{4},\frac{3\pi }{4}\}$ (2 solutions)
Total solutions for $a=0$ is $5$.

Case 2: $a=1$
${\sin }^{2}x+\sin x\cos x=1$
$\Rightarrow \sin x\cos x=1-{\sin }^{2}x$
$\Rightarrow \sin x\cos x={\cos }^{2}x$
$\Rightarrow \cos x(\sin x-\cos x)=0$
$\Rightarrow \cos x=0$ or $\sin x-\cos x=0$
For $x\in [-\pi ,\pi ]$, $\cos x=0\Rightarrow x\in \{-\frac{\pi }{2},\frac{\pi }{2}\}$ (2 solutions)
$\sin x-\cos x=0\Rightarrow \tan x=1\Rightarrow x\in \{-\frac{3\pi }{4},\frac{\pi }{4}\}$ (2 solutions)
Total solutions for $a=1$ is $4$.

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

Answer: $9$