Solution
= Solution
Since $\cos x>0$ throughout the given <interval>, multiplying by $\cos x$ is reversible. The equation simplifies to
$$
2-\sin x=\sqrt3\cos x,\qquad \sin x+\sqrt3\cos x=2\sin(x+\pi/3)=2.
$$
The <sine> equals $1$ only at $\pi/2+2j\pi$. Here $x+\pi/3\in[\pi/3,5\pi/6)$, so only $x+\pi/3=\pi/2$ is allowed. Hence
$$
\boxed{x=\pi/6.}
$$