-1
$\begingroup$

I don't know how to solve this kind of trig equation, could you help me?

$$3 \tan(\pi-x)+\tan^3(x)=0$$

  • 9
    $\tan (\pi - x) = - \tan x$. Now call $\tan x = t$ and solve for $t$. – 2017-01-02
  • 1
    @ZacharySelk's comment has at least the virtue of signalling to the OP that there is not even a question here, and that, if they could only be bothered to transform their post into a question, they might (I said, might) realize that at present its format is not adequate ("My teachers are asking me" does not count as personal input, does it?). That this message could be transmitted by other means than what Zachary wrote is true, that it should be transmitted by other means is debatable, that it should be given is, imho, not debatable. What did the other users do to convey the information? – 2017-01-02
  • 0
    You should really ask your teachers! Their purpose in this universe is precisely to explain this to you. – 2017-01-02
  • 1
    @JoanaPaulo: Welcome to Math.SE! Because this is a [Question and Answer site](http://math.stackexchange.com/tour) and not a chat/discussion site, it's preferable to make the titles of your questions concise summaries of what you're asking. The current title is not a mathematics question (and would be off-topic for this site), while the apparent content of the question concerns solving a particular trig equation (and is on-topic). Could you please edit the question body so that your mathematical question is clearer, and edit the title to reflect your mathematical question? Thanks. :) – 2017-01-02

2 Answers 2

5

You can recall that $\tan(\pi-x)=-\tan x$; so, if you set $y=\tan x$, the equation becomes $$ y^3-3y=0 $$ Can you solve it? Can you then solve $\tan x=y$ for the found values of $y$?

  • 1
    yes, thanks a lot – 2017-01-02
1

tan(π - x) = - tan x

So equation becomes,

-3 tanx + $\tan^3x$ = 0

Taking tanx common,

tanx (-3 + $\tan^2x$) = 0

When tanx = 0

tanx = tan0 or tanx = tan nπ

x = 0 or nπ

When -3 + $\tan^2x$ = 0

$\tan^2x$ = 3

tanx = $\pm\sqrt3$

Case 1-

tanx = $\tan{\frac{π}{3}}$ or tanx = $\tan{\frac{π}{3} \pm πn}$

x = $\frac{π}{3}$ or $\frac{π}{3} \pm πn$

Case 2-

tanx = -$\tan{\frac{π}{3}}$

x = $\frac{2π}{3}$ and $\frac{5π}{3}$

  • 1
    I didn't downvote, but it looks like your solution is incomplete. $\tan(x) = 0$ when $x = 0$ or $x = \pi$ (at least for primary solutions in $[0,2 \pi)$). – 2017-01-02
  • 0
    Also, $\tan(x) = \sqrt{3}$ both as $x = \pi/3$ and also $x = \frac{4\pi}{3}$, again, at least for primary solutions, i.e., solutions in $[0,2\pi)$. – 2017-01-02
  • 0
    thanks a lot for your help :) – 2017-01-02
  • 0
    Mine pleasure. :-) – 2017-01-02
  • 0
    @user46944 thank you. And I edited my answer. Hope its complete now. – 2017-01-02
  • 0
    There's another problem with your answer. When we have $\tan^{2}(x) = 3$, this gives two equations: $\tan(x) = \sqrt{3}$ or $\tan(x) = -\sqrt{3}$. Both need to be solved. You have only solved the first. For $\tan(x) = -\sqrt{3}$, the $x$ values are $x = \frac{2\pi}{3}$ and $\frac{5\pi}{3}$. – 2017-01-02
  • 1
    @user46944 thank u once again for correcting me – 2017-01-02
  • 0
    To answer your comment about (lack of) upvotes: I didn't downvote, but on passing through, notice that Enrico's answer 1. is clear and concise; 2. gives only a hint, leaving the OP with a pedagogically-constructive amount of work. Conversely (if you'll forgive a bit of criticism), this answer is not uniformly typeset using MathJaX, and (as currently written) omits infinitely many solutions, which in turn is a symptom of not carefully organizing cases and accommodating periodicity of the tangent function. All this makes your careful work a little difficult to read and understand. – 2017-01-02
  • 0
    @Andrew D. Hwang thank u. I understand your meaning. But the main problem is that I always use my phone to answer questions. Sometimes I skip few things because of this. But from the next time I keep these things in mind. Once again thank you. – 2017-01-02
  • 0
    Can you send me the link of mathjax. – 2017-01-02
  • 0
    Here's a [tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) in the meta site, and a link to the site's (brief) LaTeX [editing help](http://math.stackexchange.com/editing-help#latex). You can also get to the editing help page from the editing window of the site interface (i.e., when you're composing an answer) by clicking the question mark in the upper right corner of the text entry field. I hope that's helpful. :) – 2017-01-02
  • 0
    "But the main problem is that I always use my phone to answer questions" And maybe you should stop using your phone for this, if using it leads you to post answers systematically garbled with typos? – 2017-01-02