JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Trigonometric Equations

Writing $f(x)={\sin }^{2}x+\sin x\cos x=\frac{1+\sin 2x-\cos 2x}{2}$ and $\sin 2x-\cos 2x=\sqrt{2}\sin \left(2x-\frac{\pi }{4}\right)\in [-\sqrt{2},\sqrt{2}]$, the range of $f$ is $\left[\frac{1-\sqrt{2}}{2},\frac{1+\sqrt{2}}{2}\right]\approx [-0.21,1.21]$, so the only integer values are $a=0$ and $a=1$. For $a=0$: $\sin x=0$ gives $x=-\pi ,0,\pi$ and $\sin x+\cos x=0$ gives $x=-\frac{\pi }{4},\frac{3\pi }{4}$, i.e. $5$ solutions. For $a=1$: $\sin 2x-\cos 2x=1$ gives $\sin \left(2x-\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$, whose roots in $[-\pi ,\pi ]$ satisfying the original equation are $x=-\frac{3\pi }{4},-\frac{\pi }{2},\frac{\pi }{4},\frac{\pi }{2}$, i.e. $4$ solutions. Altogether $n(S)=5+4=9$.