JEE Main 2026 — 2 April, Morning Shift — Question 30

$\sin \mathrm{x}(\sin \mathrm{x}+\cos \mathrm{x}) \in\left[\frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right]$ 2 integer will be there $\Rightarrow \mathrm{a}=0,1$ If $\mathrm{a}=0 \sin \mathrm{x}(\sin \mathrm{x}+\cos \mathrm{x})=0$ $\Rightarrow \sin \mathrm{x}=0 \quad$ or $\sin \mathrm{x}+\cos \mathrm{x}=0$ $\mathrm{x}=-\pi, 0, \pi$ $=\tan \mathrm{x}=-1$

3 solutions

$$\mathrm{x}=-\frac{\pi}{4}, \frac{3 \pi}{4} 2 \text { solutions }$$

If $\mathrm{a}=1 \sin \mathrm{x}(\sin \mathrm{x}+\cos \mathrm{x})=1$ $\sin ^{2} \mathrm{x}+\sin \mathrm{x} \cos \mathrm{x}=1$ $2 \sin ^{2} \mathrm{x}+2 \sin \mathrm{x} \cos \mathrm{x}=2$ $1-\cos 2 \mathrm{x}+\sin 2 \mathrm{x}=2$ $\operatorname{Sin} 2 \mathrm{x}-\cos 2 \mathrm{x}=1 \rightarrow$ square $\operatorname{Sin} 4 \mathrm{x}=0$ $4 \mathrm{x}=-4 \pi,-3 \pi$, $\_\_\_\_$ $3 \pi, 4 \pi$ $\mathrm{x}=-\pi,-\frac{3 \pi}{4},-\frac{\pi}{2},-\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi$ $\Rightarrow \mathrm{x}=-\frac{3 \pi}{4},-\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{2}$ 4 solutions

Total 9 solution