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It's been 20 years since I did trig, and this one seems a little tricky. How would I solve $$ \tan^2(x) -2\tan(x)=1 $$ with steps?

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    $\begingroup$ Let $\tan x = t\,$ and solve $t^2-2t=1\,$ first. $\endgroup$
    – dxiv
    Apr 1, 2017 at 18:34
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    $\begingroup$ @MaLio. You will need to quadratic formula. Still remember that one? $\endgroup$
    – imranfat
    Apr 1, 2017 at 18:42
  • $\begingroup$ yes, however I don't think that's the way to go about solving the problem. This is in my daughters textbook in the trig identities section. (yes, I am trying to help with homework). I like haqnatural solution below, but how does the tan^2x-2tanx-> (tanx -1)^2 - 1 ? $\endgroup$
    – MaLio
    Apr 1, 2017 at 18:46
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    $\begingroup$ If she is learning trig identities, then she likely knows the quadratic formula and can use it. $\endgroup$
    – Kaynex
    Apr 1, 2017 at 18:58

3 Answers 3

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$$\tan ^{ 2 }{ x-2\tan { x } -1=0 } \\ \tan ^{ 2 }{ x-2\tan { x } +1-2=0 } \\ { \left( \tan { x } -1 \right) }^{ 2 }-2=0\\ \tan { x-1=\pm \sqrt { 2 } } \\ \tan { x } =1\pm \sqrt { 2 } \\ x=\arctan { \left( 1\pm \sqrt { 2 } \right) +\pi n } \\ $$

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  • $\begingroup$ how did you get from the first to the second line? (as in what identity did you use) $\endgroup$
    – MaLio
    Apr 1, 2017 at 18:55
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    $\begingroup$ @MaLio Completing the Square. $\endgroup$ Apr 1, 2017 at 19:00
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    $\begingroup$ That isn't a trig identity, but instead an application of completing the square. $x^2 - 2x + 1 = (x - 1)^2$ $\endgroup$
    – Kaynex
    Apr 1, 2017 at 19:00
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Put $\tan x = z$

Then equation becomes,

$z^2 - 2z = 1$

$z^2 - 2z - 1 = 0$

Hope you can now factorise.

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let $B = \tan(x)$

$\tan(x)$ varies with $x$, but ultimately is just a value

now rewriting the equation to give $B^2 -2B -1 = 0$, This will not factorise with integers, but solves to give $x = 1$ plus or minus square root $2$

from there, we use the $\arctan$ function and it tells you that (in radians) x = $\arctan(1 + \sqrt2)$ or $\arctan(1 - \sqrt2$) which will give you two roots every $2\pi$ radians, so you need to restrict the range of the function to get any real answers

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