Q.Solve $\sqrt3\tan\theta-1=0$ for the general solution, and hence list every solution with $0\le\theta<2\pi$.

Once $\tan\theta$ is isolated and matched to a standard angle, the tangent general-solution formula (which has period $\pi$, not $2\pi$) generates every solution, and substituting small integer values of $n$ lists the ones inside one full rotation.

Step 1 — isolate the ratio. $\sqrt3\tan\theta-1=0 \Rightarrow \tan\theta=\dfrac{1}{\sqrt3}$.

Step 2 — identify the standard angle. $\dfrac{1}{\sqrt3}=\tan30^\circ=\tan\dfrac{\pi}{6}$, so $\alpha=\dfrac{\pi}{6}$.

Step 3 — apply the general-solution formula for tangent. $\tan\theta=\tan\alpha \Rightarrow \theta=n\pi+\alpha$, giving $\theta=n\pi+\dfrac{\pi}{6}$, $n\in\mathbb{Z}$.

Step 4 — list solutions in $[0,2\pi)$. $n=0$: $\theta=\dfrac{\pi}{6}$. $n=1$: $\theta=\pi+\dfrac{\pi}{6}=\dfrac{7\pi}{6}$. $n=2$ gives $\theta=2\pi+\dfrac{\pi}{6}$, which is outside the range, so only these two values qualify. …

Using the sine/cosine general-solution formula (period $2\pi$, with the $(-1)^n$ or $\pm$ pattern) instead of the tangent formula (period $\pi$, no sign alternation) — this is the single most common confusion in thi …