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How do you solve the following trigonometric equation? $$\tan x - 1 = -\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} \cot x.$$

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up vote 6 down vote accepted

(You solve an equation, not a function.)

Perhaps the easiest approach is to let $u=\tan x$ and use the fact that $\cot x=\frac1{\tan x}$ to rewrite it as $$u-1=-\frac1{\sqrt3}+\frac1{\sqrt3u}=\frac{1-u}{\sqrt3u}\;.$$ Now solve for $u$; at that point you'll have $\tan x$, and the rest is plain sailing.

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We have $\displaystyle \cot(x) = \frac1{\tan(x)}$. Hence, the equation gives us $\displaystyle \tan(x) - 1 = - \frac1{\sqrt{3}} + \frac1{\sqrt{3}} \frac1{\tan(x)}$. Multiplying by $\tan(x)$, we get that $$\tan^2(x) - \tan(x) = - \frac{\tan(x)}{\sqrt{3}} + \frac1{\sqrt{3}}$$ Setting $\displaystyle m = \tan(x)$, we get the quadratic, $$m^2 + \left( \frac1{\sqrt{3}} - 1\right)m - \frac1{\sqrt{3}} = 0$$ which gives us that $$m = \frac{-\left( \frac1{\sqrt{3}} - 1\right) \pm \sqrt{\left(\frac1{\sqrt{3}} - 1\right)^2 + \frac4{\sqrt{3}}}}{2} = \frac{-\left( \frac1{\sqrt{3}} - 1\right) \pm \left(\frac1{\sqrt{3}} + 1\right)}{2} = -\frac1{\sqrt{3}}, 1$$

$\displaystyle m = 1$ gives us $\displaystyle x = n \pi + \frac{\pi}{4}$ while $\displaystyle m = - \frac1{\sqrt{3}}$ gives us $\displaystyle x = n \pi - \frac{\pi}{6}$ where $n \in \mathbb{Z}$.

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5  
A hint or incomplete solution would have been preferable, since this is homework. – Brian M. Scott May 7 '12 at 1:13
5  
@BrianM.Scott I used to give hints. But I have now become indifferent to homework problem. – user17762 May 7 '12 at 1:14
3  
@Brian: Blunty, you should probably just accept it. I fought the "pedagogical" style of spoon-feeding for a long time... maybe I should still be in the fight, but it seems we're outnumbered! – The Chaz 2.0 May 7 '12 at 1:25
3  
@Brian: Of course! I should probably just accept that some people don't remain silent about things that they find unacceptable... or maybe I should just remain silent :) – The Chaz 2.0 May 7 '12 at 1:35
7  
@Marvis: Bluntly, I consider that indefensible. It ill serves the people asking the questions (and their teachers, for that matter). – Brian M. Scott May 7 '12 at 1:42

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