Question

4tan^2 x-8tanx+3=0 on interval [0,2pi]

Answer 1

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