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![\bf 3tan(2x)=-3\implies tan(2x)=\cfrac{-3}{3}\implies tan(2x)=-1 \\\\\\ tan^{-1}[tan(2x)]=tan^{-1}(-1)\implies 2x=tan^{-1}(-1) \\\\\\ 2x= \begin{cases} \frac{3\pi }{4}\\\\ \frac{7\pi }{4} \end{cases}\implies \begin{cases} 2x=\cfrac{3\pi }{4}\implies \measuredangle x=\cfrac{3\pi }{8}\\\\ 2x=\cfrac{7\pi }{4}\implies \measuredangle x=\cfrac{7\pi }{8} \end{cases}](/tpl/images/0325/7590/2e89b.png)
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Solve 3 tan 2x = -3 in the interval [0, 2π)...
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