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Thread: Solving Trig equations using the "t" substitution

  1. #1
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    Solving Trig equations using the "t" substitution

    Hi folks,


    Normally when solving trig equations, for example: $tan \ x = -\frac{1}{3}$, then $tan^{-1} (\frac{1}{3}) = 18.4$ then because the tangent is negative, solutions are found in the second and fourth quadrants. So given a range $0 \le x \le 360 $ the solutions are 161.6 degrees and 341.6 degrees.


    Now, I am trying to solve a trig equation using the $t = tan(\frac{x}{2})$ substitution. In this:


    $ tan \ x = \dfrac{2t}{1 - t^2}, \ sin \ x = \dfrac{2t}{1 + t^2}, \ cos \ x = \dfrac{1 - t^2}{1 + t^2} $


    So, I am trying to solve:


    $7 \ cos \ x + 6 \ sin \ x = 2$


    and using the substitution, I get $t = -\frac{1}{3}$ and $t = \frac{5}{3}$.


    therefore


    $\frac{x}{2} = - 18.4$ degrees $\Rightarrow x = - 36.9 $ degrees


    As mentioned above, the solutions are in quad 2 and 4 so for the given range we should get x = 143.1 and x = 323.1


    My maths books only gives x = 323.1 as a solution. For some reason, the solution in the second quad is ignored. Why would this be?


    For $tan \frac{x}{2} = \frac{5}{3}$ we get x = 118.1 and this is again the only solution provided. I would expect a solution in quad 4 (298.1).


    I am wondering whether the fact that $\frac{x}{2} = 59$ degrees is in quad 1 and x = 118.1 is in quad 2 is the reason.


    Can anyone explain this?
    Last edited by s_ingram; Aug 6th 2017 at 05:10 AM.
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  2. #2
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    Re: Solving Trig equations using the "t" substitution

    note ...

    $0 \le x < 360^\circ \implies 0\le \dfrac{x}{2} < 180^\circ$

    ... so when using the $t$ substitution, $\dfrac{x}{2}$ will be calculated only for quads I and II



    since $t = \tan\left(\dfrac{x}{2}\right) = -\dfrac{1}{3} < 0 \implies \dfrac{x}{2} = \tan^{-1}\left(-\dfrac{1}{3}\right) \implies 90^\circ < \dfrac{x}{2} < 180^\circ$

    $\dfrac{x}{2} = 180 + \tan^{-1}\left(-\dfrac{1}{3}\right) \approx 161.6^\circ \implies x \approx 323.1^\circ$


    $\dfrac{x}{2} = \tan^{-1}\left(\dfrac{5}{3}\right) \implies 0^\circ < \dfrac{x}{2} < 90^\circ$

    $\dfrac{x}{2} = \tan^{-1}\left(\dfrac{5}{3}\right) \approx 59.04^\circ \implies x \approx 118.1^\circ$
    Thanks from s_ingram
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    Re: Solving Trig equations using the "t" substitution

    Hi skeeter,

    many thanks once again. It is amazing how many ways there are of getting questions wrong!
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