Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 8 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.

share|cite|improve this answer

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 $n \in \mathbb{Z}$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.