[tex]\bf 4tan^2(x)+8tan(x)+3=0\qquad [0,2\pi ]
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\textit{since it's just a quadratic, we'll just factor it}
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[2tan(x)+1][2tan(x)+3]=0\implies
\begin{cases}
2tan(x)+1=0\\
\qquad tan(x)=-\frac{1}{2}\\
\qquad \measuredangle x=tan^{-1}\left( -\frac{1}{2} \right)\\
\qquad \measuredangle x\approx 333.4^o, 153.4^o\\
2tan(x)+3=0\\
\qquad tan(x)=-\frac{3}{2}\\
\qquad \measuredangle x=tan^{-1}\left(-\frac{3}{2} \right)\\
\qquad \measuredangle x\approx303.7^o, 123.7^o
\end{cases}[/tex]
