Lucia Grimes

2022-07-07

solve:
$\sqrt{3}\mathrm{cot}A.\mathrm{cot}2A-\mathrm{cot}A-\mathrm{cot}2A=\sqrt{3}$

conveneau71

Expert

$\sqrt{3}\mathrm{cot}A.\mathrm{cot}2A-\mathrm{cot}A-\mathrm{cot}2A-\sqrt{3}=0$
$\sqrt{3}\left(\mathrm{cot}A\cdot \mathrm{cot}2A-1\right)-1\left(\mathrm{cot}A+\mathrm{cot}2A\right)=0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\sqrt{3}\left(\mathrm{cot}A\cdot \mathrm{cot}2A-1\right)=\left(\mathrm{cot}A+\mathrm{cot}2A\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\sqrt{3}=\frac{\mathrm{cot}A+\mathrm{cot}2A}{\mathrm{cot}A\cdot \mathrm{cot}2A-1}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\sqrt{3}=\frac{\mathrm{tan}2A+\mathrm{tan}A}{1-\mathrm{tan}A\cdot \mathrm{tan}2A}=\mathrm{tan}\left(2A+A\right)=\mathrm{tan}3A$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{tan}3A=\sqrt{3}=\mathrm{tan}\frac{\pi }{3}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}3A=n\pi +\frac{\pi }{3}$
So $A=\frac{n\pi }{3}+\frac{\pi }{9}$ where n=0,1,2,…
Now choose n such that ${0}^{\circ } according to your question.

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