determine the roots of the equation 7 sec 2 </msup> &#x2061;<!-- ⁡ --> <mrow cla

determine the roots of the equation $7{\mathrm{sec}}^{2}x+2\mathrm{tan}x-6=2{\mathrm{sec}}^{2}x+2$ when $0\le x\le 2\pi$
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Ordettyreomqu
Using $1+ta{n}^{2}=se{c}^{2}$,
$7{\mathrm{sec}}^{2}\left(x\right)+2\mathrm{tan}\left(x\right)-6=2{\mathrm{sec}}^{2}\left(x\right)+2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$7+7{\mathrm{tan}}^{2}\left(x\right)+2\mathrm{tan}\left(x\right)-6=2+2{\mathrm{tan}}^{2}\left(x\right)+2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$5{\mathrm{tan}}^{2}\left(x\right)+2\mathrm{tan}\left(x\right)-3=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\left(\mathrm{tan}\left(x\right)+1\right)\left(5\mathrm{tan}\left(x\right)-3\right)=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$

Thus, in $\left[0,2\pi \right)$, $x=3\pi /4$ or $x=7\pi /4$ or $x={\mathrm{tan}}^{-1}\left(3/5\right)$ or $x={\mathrm{tan}}^{-1}\left(3/5\right)+\pi .$