Question

Find all solutions of the equation on the interval [0,2π]

Answer 1

1)

$\bf 3tan(x)-\sqrt{3}=0\implies 3tan(x)=\sqrt{3}\implies \boxed{tan(x)=\cfrac{\sqrt{3}}{3}}$

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.

$\bf \cfrac{\pi }{6}\quad \begin{cases} sine=&\frac{1}{2}\\ cosine=&\frac{\sqrt{3}}{2} \end{cases}\qquad tan\left( \frac{\pi }{6}\right)=\cfrac{sin\left( \frac{\pi }{6}\right)}{cos\left( \frac{\pi }{6}\right)}\implies \cfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$\bf \cfrac{1}{2}\cdot \cfrac{2}{\sqrt{3}}\implies \cfrac{1}{\sqrt{3}}\impliedby \textit{now, let's \underline{rationalize the denominator}} \\\\\\ \cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{1\sqrt{3}}{(\sqrt{3})^2}\implies \boxed{\cfrac{\sqrt{3}}{3}}$

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.

$\bf \measuredangle x= \begin{cases} \frac{\pi }{6}&I~Quadrant\\\\ \frac{7\pi }{6}&III~Quadrant \end{cases}$



2)

$\bf 4cos^2(x)-1=0\implies 4cos^2(x)=1\implies cos^2(x)=\cfrac{1}{4} \\\\\\ cos(x)=\pm\sqrt{\cfrac{1}{4}}\implies cos(x)=\pm\cfrac{\sqrt{1}}{\sqrt{4}}\implies cos(x)=\pm\cfrac{1}{2} \\\\\\ \measuredangle x=cos^{-1}\left( \pm\frac{1}{2} \right)\implies \measuredangle x= \begin{cases} \frac{\pi }{3}\\\\ \frac{2\pi }{3}\\\\ \frac{4\pi }{3}\\\\ \frac{5\pi }{3} \end{cases}$