1. We are given the equation $\sqrt{3} \sin(x) - \cos(x) = 0$ and need to find the values of $x$ that satisfy it. 2. Rewrite the equation to isolate terms: $\sqrt{3} \sin(x) = \cos(x)$. 3. Divide both sides by $\cos(x)$ (where $\cos(x) \neq 0$): $$ \sqrt{3} \frac{\sin(x)}{\cos(x)} = 1 $$ 4. Use the identity $\tan(x) = \frac{\sin(x)}{\cos(x)}$: $$ \sqrt{3} \tan(x) = 1 $$ 5. Solve for $\tan(x)$: $$ \tan(x) = \frac{1}{\sqrt{3}} $$ 6. Recall that $\tan \theta = \frac{1}{\sqrt{3}}$ at angles $\theta = \frac{\pi}{6} + k\pi$, for any integer $k$. 7. Therefore, the solutions for $x$ are: $$ x = \frac{\pi}{6} + k\pi, \quad k \in \mathbb{Z} $$ 8. This means $x$ equals $\frac{\pi}{6}$ plus any integer multiple of $\pi$ radians. Final Answer: $$ x = \frac{\pi}{6} + k\pi, \quad k \in \mathbb{Z} $$