Gybrisysmemiau7

Answered

2022-06-26

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

Answer & Explanation

last99erib

Expert

2022-06-27Added 19 answers

Squaring both sides is false way because in this case you can get false roots

$1+\sqrt{3}\mathrm{tan}\alpha -\mathrm{sec}\alpha =0\phantom{\rule{0ex}{0ex}}1+\sqrt{3}\frac{\mathrm{sin}\alpha}{\mathrm{cos}\alpha}-\frac{1}{\mathrm{cos}\alpha}=0\phantom{\rule{0ex}{0ex}}\mathrm{cos}\alpha +\sqrt{3}\mathrm{sin}\alpha =1\phantom{\rule{0ex}{0ex}}\frac{1}{2}\mathrm{cos}\alpha +\frac{\sqrt{3}}{2}\mathrm{sin}\alpha =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}{30}^{\circ}\mathrm{cos}\alpha +\mathrm{cos}{30}^{\circ}\mathrm{sin}\alpha =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}({30}^{\circ}+\alpha )=\frac{1}{2}\phantom{\rule{0ex}{0ex}}{30}^{\circ}+\alpha ={30}^{\circ}+{360}^{\circ}n,\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{\alpha}_{1}={360}^{\circ}n,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}n=0,\pm 1,\pm 2,...\phantom{\rule{0ex}{0ex}}{30}^{\circ}+\alpha ={180}^{\circ}-{30}^{\circ}+{360}^{\circ}n\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{\alpha}_{2}={120}^{\circ}+{360}^{\circ}n,\phantom{\rule{1em}{0ex}}n=0,\pm 1,\pm 2,.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$1+\sqrt{3}\mathrm{tan}\alpha -\mathrm{sec}\alpha =0\phantom{\rule{0ex}{0ex}}1+\sqrt{3}\frac{\mathrm{sin}\alpha}{\mathrm{cos}\alpha}-\frac{1}{\mathrm{cos}\alpha}=0\phantom{\rule{0ex}{0ex}}\mathrm{cos}\alpha +\sqrt{3}\mathrm{sin}\alpha =1\phantom{\rule{0ex}{0ex}}\frac{1}{2}\mathrm{cos}\alpha +\frac{\sqrt{3}}{2}\mathrm{sin}\alpha =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}{30}^{\circ}\mathrm{cos}\alpha +\mathrm{cos}{30}^{\circ}\mathrm{sin}\alpha =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\mathrm{sin}({30}^{\circ}+\alpha )=\frac{1}{2}\phantom{\rule{0ex}{0ex}}{30}^{\circ}+\alpha ={30}^{\circ}+{360}^{\circ}n,\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{\alpha}_{1}={360}^{\circ}n,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}n=0,\pm 1,\pm 2,...\phantom{\rule{0ex}{0ex}}{30}^{\circ}+\alpha ={180}^{\circ}-{30}^{\circ}+{360}^{\circ}n\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{\alpha}_{2}={120}^{\circ}+{360}^{\circ}n,\phantom{\rule{1em}{0ex}}n=0,\pm 1,\pm 2,.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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