Respuesta :

using the range of [ 0 , 2π ]


[tex]\bf f(x)=2sec(x)+tan(x)\implies \cfrac{df}{dx}=2sec(x)tan(x)+sec^2(x) \\\\\\ 0=2sec(x)tan(x)+sec^2(x) \\\\\\ -sec^2(x)=2sec(x)tan(x)\implies \cfrac{-sec^2(x)}{2sec(x)}=tan(x) \\\\\\ \cfrac{-sec(x)}{2}=tan(x)\implies \cfrac{-\frac{1}{cos(x)}}{\frac{2}{1}}=\cfrac{sin(x)}{cos(x)} \\\\\\ -\cfrac{1}{2cos(x)}=\cfrac{sin(x)}{cos(x)}\implies -\cfrac{1}{2}=sin(x)\implies \measuredangle x = \begin{cases} \frac{7\pi }{6}\\\\ \frac{11\pi }{6} \end{cases}[/tex]

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