Gybrisysmemiau7

2022-06-26

Solve for $\alpha$ where ${0}^{\circ }\le \alpha \le {360}^{\circ }$, in $1+\sqrt{3}\mathrm{tan}\alpha -\mathrm{sec}\alpha =0$

last99erib

Expert

Squaring both sides is false way because in this case you can get false roots
$1+\sqrt{3}\mathrm{tan}\alpha -\mathrm{sec}\alpha =0\phantom{\rule{0ex}{0ex}}1+\sqrt{3}\frac{\mathrm{sin}\alpha }{\mathrm{cos}\alpha }-\frac{1}{\mathrm{cos}\alpha }=0\phantom{\rule{0ex}{0ex}}\mathrm{cos}\alpha +\sqrt{3}\mathrm{sin}\alpha =1\phantom{\rule{0ex}{0ex}}\frac{1}{2}\mathrm{cos}\alpha +\frac{\sqrt{3}}{2}\mathrm{sin}\alpha =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}{30}^{\circ }\mathrm{cos}\alpha +\mathrm{cos}{30}^{\circ }\mathrm{sin}\alpha =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}\left({30}^{\circ }+\alpha \right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}{30}^{\circ }+\alpha ={30}^{\circ }+{360}^{\circ }n,\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{\alpha }_{1}={360}^{\circ }n,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}n=0,±1,±2,...\phantom{\rule{0ex}{0ex}}{30}^{\circ }+\alpha ={180}^{\circ }-{30}^{\circ }+{360}^{\circ }n\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{\alpha }_{2}={120}^{\circ }+{360}^{\circ }n,\phantom{\rule{1em}{0ex}}n=0,±1,±2,.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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