1)
[tex]\bf 3tan(x)-\sqrt{3}=0\implies 3tan(x)=\sqrt{3}\implies \boxed{tan(x)=\cfrac{\sqrt{3}}{3}}[/tex]
now, let's take a peek at your Unit Circle, as you should have one, if you don't, this is a prime time to get one, you'll need it, you can find many online, or I can post one here for you, many good ones. Anyhow, let's take a peek at π/6.
[tex]\bf \cfrac{\pi }{6}\quad
\begin{cases}
sine=&\frac{1}{2}\\
cosine=&\frac{\sqrt{3}}{2}
\end{cases}\qquad tan\left( \frac{\pi }{6}\right)=\cfrac{sin\left( \frac{\pi }{6}\right)}{cos\left( \frac{\pi }{6}\right)}\implies \cfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}[/tex]
[tex]\bf \cfrac{1}{2}\cdot \cfrac{2}{\sqrt{3}}\implies \cfrac{1}{\sqrt{3}}\impliedby \textit{now, let's \underline{rationalize the denominator}}
\\\\\\
\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{1\sqrt{3}}{(\sqrt{3})^2}\implies \boxed{\cfrac{\sqrt{3}}{3}}[/tex]
now, the tangent function is positive, if both the numerator and denominator have the same sign, that happens in the I Quadrant, as in π/6, but is also true in the III Quadrant, since both are negative.
[tex]\bf \measuredangle x=
\begin{cases}
\frac{\pi }{6}&I~Quadrant\\\\
\frac{7\pi }{6}&III~Quadrant
\end{cases}[/tex]
2)
[tex]\bf 4cos^2(x)-1=0\implies 4cos^2(x)=1\implies cos^2(x)=\cfrac{1}{4}
\\\\\\
cos(x)=\pm\sqrt{\cfrac{1}{4}}\implies cos(x)=\pm\cfrac{\sqrt{1}}{\sqrt{4}}\implies cos(x)=\pm\cfrac{1}{2}
\\\\\\
\measuredangle x=cos^{-1}\left( \pm\frac{1}{2} \right)\implies \measuredangle x=
\begin{cases}
\frac{\pi }{3}\\\\
\frac{2\pi }{3}\\\\
\frac{4\pi }{3}\\\\
\frac{5\pi }{3}
\end{cases}[/tex]
