Trigonometric Equations

The key technique is to convert the linear combination $a\sin x+b\cos x$ into a single sinusoid with a phase shift, since this reduces an awkward two-term equation to a basic one whose solutions can be counted directly. Factoring out the amplitude $\sqrt{1+3}=2$, we write $\sin x+\sqrt{3}\cos x=2\left(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x\right)=2\sin \left(x+\frac{\pi }{3}\right)$, recognizing the coefficients as $\cos \frac{\pi }{3}$ and $\sin \frac{\pi }{3}$. The equation now reads $2\sin \left(x+\frac{\pi }{3}\right)=1$, so $\sin \left(x+\frac{\pi }{3}\right)=\frac{1}{2}$. The interval matters: as $x$ ranges over the open interval $(0,\pi )$, the shifted argument $x+\frac{\pi }{3}$ ranges over $\left(\frac{\pi }{3},\frac{4\pi }{3}\right)$. Within this transformed interval, $\sin$ equals $\frac{1}{2}$ only at $\frac{5\pi }{6}$, because the other usual solution $\frac{\pi }{6}$ falls below the interval's lower bound. This gives the single value $x=\frac{5\pi }{6}-\frac{\pi }{3}=\frac{\pi }{2}$. Option 2 wrongly counts $\frac{\pi }{6}$, which lies outside the shifted range. Option 0 overlooks the valid root entirely. Option 3 over-counts by including periodic images beyond the interval. This follows the standard amplitude-conversion equation pattern of JEE Advanced. As a final check, substituting $x=\frac{\pi }{2}$ into the original gives $1+0=1$, confirming exactly one solution.