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How to solve this trig equation

  1. Jul 21, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex] \tan(2x)-\cot(3x)=0 [/tex], where, [tex] 0 \le x < 2\pi [/tex]
    I have tried to plot the figure of this equation, but it seems so complicated.


    2. Relevant equations



    3. The attempt at a solution
    Using the double-angle formula [tex]\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}[/tex] and
    identity [tex]\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}[/tex], we get

    [tex]\tan(3x)=\tan(x+2x)=\frac{\tan(x)+\tan(2x)}{1-\tan(x)\tan(2x)}[/tex],

    and then

    [tex]\cot(3x)=\frac{1}{\tan(x+2x)}=\frac{1-\tan(x)\tan(2x)}{\tan(x)+\tan(2x)}=\frac{1-\tan(x)\frac{2\tan(x)}{1-\tan^2(x)}}{\tan(x)+\frac{2\tan(x)}{1-\tan^2(x)}}[/tex]

    After some simplification, we have

    [tex]\cot(3x)=\frac{1-3\tan^2(x)}{3\tan(x)-\tan^3(x)}[/tex]

    Adding our results [tex]\tan(2x)[/tex] and [tex]\cot(3x)[/tex],
    the equation can be rewritten as:

    [tex]5\tan(x)^4-10\tan(x)^2+1=0[/tex]

    Therefore I get:

    [tex]\tan(x)=\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}[/tex]

    and

    [tex]x=\pm\arctan\left(\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right)[/tex]

    But it seems that my solution is not completed, some solutions are missing.

    As Mentallic mentioned, it is not appropriate to drag out the plus minus sign in the arctan function, so the solutions should be written as

    [tex]x=\arctan\left(\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right),\pi+\arctan\left(\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right)[/tex]

    However, we find that two of the eight solutions are negative, where the problem requires [tex]0 \le x < 2\pi[/tex].

    These two solutions are [tex]\arctan\left(-\sqrt{1+\frac{2}{5}\sqrt{5}}\right)[/tex] and [tex]\arctan\left(-\sqrt{1-\frac{2}{5}\sqrt{5}}\right)[/tex]

    Since for a given value [tex]y=\tan(x)[/tex], it should correspond to two angle [tex]x[/tex] for a given range [tex]0 \le x < 2\pi[/tex]. In this case, we simply add [tex]2\pi[/tex] to those two negative solutions, the correct solutions can be obtained.

    Finally, we get the results:

    [tex]x=\arctan\left(\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right),\pi+\arctan\left(\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right),2\pi+\arctan\left(-\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right)[/tex]

    However, as indicated by Mentallic at #4, we are still missing two solutions due to the fact that we simplified the equation by multiplying [tex](1-\tan^2(x))(3\tan(x)-\tan^3(x))[/tex] which is equivalent to divide the equation by [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}[/tex]. In such case it implies that we assumed [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}\neq 0[/tex] that in turn we lose the solutions of [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}=0[/tex]. As a result, we need to solve equation [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}=0[/tex] to find out the missing solutions.

    Equation [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}=0[/tex] can be rewritten as

    [tex]\frac{1}{3\tan(x)-4\tan^3(x)+\tan^5(x)}=0[/tex]

    That means the denominator is undefined. For [tex]\tan(x),\tan^3(x),\tan^5(x)[/tex], they have the same undefined points, which are [tex]\frac{\pi}{2},\frac{3\pi}{2}[/tex], therefore their combination [tex]3\tan(x)-4\tan^3(x)+\tan^5(x)[/tex]. In conclusion, the two missing solutions are [tex]x=\frac{\pi}{2},\frac{3\pi}{2}[/tex].
     
    Last edited: Jul 22, 2009
  2. jcsd
  3. Jul 21, 2009 #2

    Mentallic

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    I'm very glad you were able to apply the double-angle formulas correctly. It saves us a lot of time :smile:

    ok, so as you can see all the 4 solutions for [tex]tan(x)[/tex] are real.

    Now lets go back to the basics of trig for a second:

    For [tex]tan(\theta)=x[/tex]

    [tex]\theta=arctan(x),\pi + arctan(x)[/tex] for [tex]0\leq \theta \leq 2\pi[/tex]

    Notice that this is true for all real x. Positive OR negative.

    e.g. if x=1, then [tex]\theta=arctan(1),pi+arctan(1)=\frac{\pi}{4},\pi+\frac{\pi}{4}[/tex]

    But if x=-1, then [tex]\theta=arctan(-1)...[/tex] and [tex]arctan(-1)=-\frac{\pi}{4}[/tex]
    and the second angle will be [tex]\pi+arctan(-1)=\pi-\frac{\pi}{4}[/tex] which is in the 2nd quadrant, which should be true because that is where negative values of [tex]tan(x)[/tex] should be!


    Lets now treat this question in EXACTLY the same way.

    [tex]\tan(x)=\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}[/tex]

    [tex]x=arctan\left(\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right) , \pi + arctan\left(\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right)[/tex]

    Do not drag out the first [tex]\pm[/tex] because it is not always true. Actually, by thinking about it, it is rarely true.

    So you have 8 values of x now, but there are 10 values. Where are the other 2?
    *thinks about playing where's wally*

    Anyway, before I go any further I'd like to know exactly how you started by solving in the beginning of the question - before all the algebra - to result with the quartic. Did you instantly use the double angle formulas? Did you multiply/divide by anything? etc.
     
  4. Jul 22, 2009 #3
    Mentallic, thank you for your instruction. I have written down all the steps of my solutions. Am I correct this time? I find that two of the regular solutions are negative, which are not located in the given range. But I think that it is pretty difficult to figure out these two solutions are negative, since negative values can correspond to positive angles in tan function.
     
  5. Jul 22, 2009 #4

    Mentallic

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    I have to commend you on a job well done using the tex tags. Very nicely done! :smile:

    Yes you are correct thus far, and good job on noticing those negative values. You've already exceeded my expectations as I was at most hoping that you would notice them in the end when calculating the approximate solutions. Again, very nice!

    ok as for the last piece to the puzzle. The last 2 solutions that are missing. The problem lies in a similar principle to this:

    If you are given [tex]x^2-x=0[/tex]

    Then through factorization you will have [tex]x=0,1[/tex]

    But lets say you instead happen to divide through by x. Now you instead have [tex]x-1=0[/tex] and only the solution [tex]x=1[/tex].
    You "lost" the solution [tex]x=0[/tex] because you divided the equation through by x and thus it is assumed that [tex]x\neq 0[/tex]


    Lets go back to the question - to the algebra manipulation after the double angle formulas are used and spot the critical step that caused us to lose these two solutions:

    [tex]tan(2x)-cot(3x)=0[/tex]


    [tex]\frac{2\tan(x)}{1-\tan^2(x)} - \frac{1-3\tan^2(x)}{3\tan(x)-\tan^3(x)}=0[/tex]

    Now you would have multiplied through by the denominators: [tex](1-\tan^2(x))(3\tan(x)-\tan^3(x))[/tex]

    And finally after some more simplification you would obtain the quartic.

    The problem is when you multiplied through by the denominators... Notice how this is similar to the e.g. I gave earlier because multiplying by some variable [tex]y[/tex] is the same as dividing by [tex]\frac{1}{y}[/tex] and remember that we now assume that we have not divided by 0. We should have instead "factorized" the denominator in a sense so as not to lose the solutions.

    So what if [tex]\frac{1}{y}=0[/tex] ? Then we have lost one or more solutions.

    This is kind of an odd situation though, because now we have to solve [tex]\frac{1}{y}=0[/tex] thus [tex]y=\frac{1}{0}[/tex] which means we are searching for where the function y is undefined. Of course this isn't a problem for the tangent function though. There are plenty of undefined points :smile:

    See if you can find them.
     
  6. Jul 22, 2009 #5
    we knw tan(A+B)=tanA+ tanB/ 1-tanAtanB
    hence
    tanAtanB-1= -(tan A+tanB/tan(A+B))

    tan2xtan3x-1/tan3x= -tan2x -tan3x / tan5x
    hence solutions can be generated from
    tan2x=-tan3x. hope u knw the general formula for the solns of that.
    and rejecting all solns where tan5x=0.
    :)
     
  7. Jul 22, 2009 #6

    Mentallic

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    I don't understand how you concluded that solutions to the OP's question will be generated from [tex]tan(2x)=-tan(3x)[/tex]. And even so, this isn't true.
    Also, as for your line: " tan2xtan3x-1/tan3x= -tan2x -tan3x / tan5x "

    Please attempt to at least use brackets when required, and where did the 1/tan3x come from on the LHS?
     
  8. Jul 22, 2009 #7
    Thank you for your explanation, Mentallic, it is so impressive and clear that I never think of. I provide the last two solutions now, are they correct?
     
  9. Jul 22, 2009 #8

    Mentallic

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    heh, thanks :redface:

    That must be a rhetorical question, because I know there's no doubt in your mind that the last two solutions are correct :wink:

    This is just as a little addition, and I'm by no means saying you were wrong at all. To solve the last equation for the undefined points (which you have re-written 4 times :tongue:) I would have gone about it like this:

    [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}[/tex]

    taking the recriprocals,

    [tex](1-\tan^2(x))(3\tan(x)-\tan^3(x))=\frac{1}{0}[/tex]

    factorizing the LHS completely,

    [tex]tan(x)(1-tan(x))(1+tan(x))(\sqrt{3}-tan(x))(\sqrt{3}+tan(x))=\frac{1}{0}[/tex]

    Now there are 5 linear factors in [tex]tan(x)[/tex]
    The only possible way that the multiplication of all these factors will be undefined is if the majority of factors are undefined, while the minority are zero for the same value of x.

    Now I solve each factor equal to the undefined value, but quickly see that every factor will be reduced to [tex]tan(x)=\frac{1}{0}[/tex] because the constants will not affect the outcome.

    i.e. [tex]tan(x)+1=\frac{1}{0}[/tex] now if you minus 1 from each side, the RHS is still undefined. (In a way, I treat it as though it is infinite)

    Anyway, since all factors are the same equation, there is no need to test the solutions of x to see if there are a majority of undefined factors to zero factors because every factor will now be undefined for those values of x.

    I know this is probably hard to understand as I'm struggling to understand my own words, so here is another e.g. :smile:

    If we're given the equation [tex]\frac{x}{sin(x)}=0[/tex]

    Now lets split up the equation into factors: [tex]x(cosec(x))=0[/tex]

    So now either
    [tex]x=0[/tex]
    or
    [tex]cosec(x)=0[/tex]
    but [tex]cosec(x)\neq 0[/tex] for all real x

    Therefore we only have the solution x=0

    But if we test x=0 in the original equation: [tex](0)(cosec(0))=0\frac{1}{0} \neq 0[/tex]

    Notice how since we didn't have a majority of zero factors to undefined factors, the solution x=0 was scrapped. Actually, since we had an equal number of zero-undefined factors, [tex]f(0)[/tex] was instead a finite, non-zero value - in this case 1.

    If we tried to solve [tex]x^2cosec(x)=0[/tex] then there will be 2-zero factors while only 1-undefined factor for the value x=0, thus this equation does have the solution x=0.

    Ok I think I've rambled on enough now, I think you get the point :smile:

    Does it still seem so complicated? :biggrin:
    But I have to admit this trig question was much more complicated than I've ever been given in class. Actually I know for a fact I wouldn't have tried searching for those last two solutions in a test, because only graphing the function and counting the number of solutions made me realize I was missing them.
    Answering this question perfectly with no computer programs in test conditions would be quite an impressive feat.
     
  10. Jul 22, 2009 #9
    Again, Mentallic, thank your for your detailed explanation. In fact, this is a university entrance exam question of one of the China cities. Although calculator can be used, time is severely limited, students probably need to solve this question within five to ten minutes.
     
  11. Jul 22, 2009 #10

    Mentallic

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    Do you mean graphing calculators also? Because the only thing a regular calculator can be useful in a question like this is to give the answer as approximate solutions and to quickly confirm your results are correct... It doesn't tell you anything about how many solutions there should be.
    Don't be fooled by the 24hrs it took to finally have all the answers on this question posted in this thread. In a test it would probably take half that time! :wink:

    It has been a pleasure helping you out frensel. Good luck with the results on that exam of yours :smile:
     
  12. Jul 23, 2009 #11
    Students only can use regular calculator which cannot plot and save any mathematical formulas. I am a science teacher in fact, and in this summer, as a tutor of a friend who will take the university entrance exam next year. But I was stuck by this trig question, it is a shame indeed.
     
  13. Jul 23, 2009 #12

    Mentallic

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    Ahh I see now. Well it would be beneficial to your student if you stress how important it is to check the final solutions are valid in the original question and if any solutions are missing like has happened here.

    No shame at all, like I said, this question was tricky and unless one is very fluent with solving equations (i.e. having a good understanding of making simple manipulations and how they affect the outcome of the equation) it is more than likely mistakes will be made.
     
  14. Jul 24, 2009 #13
    When i substitute pi/2 into the original equation, cot 3x will be undefined. Can we take pi/2 as a solution?
     
  15. Jul 24, 2009 #14

    Mentallic

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    No, [tex]cot(3\pi/2)[/tex] is not undefined, it is zero. Maybe you have mistaken the cotangent function for the tangent function?
    Anyway, the solutions [tex]x=\pi/2, 3\pi/2[/tex] are solutions because [tex]tan(2x)[/tex] as well as [tex]cot(3x)[/tex] are zero for both solutions of x.
     
  16. Jul 24, 2009 #15
    i don't think it is right to say that 1/undefined = 0.
     
  17. Jul 24, 2009 #16
    yes. you're right. cot (3pi/2) is not undefined. it is 1/undefined.
     
  18. Jul 24, 2009 #17
    sorry. you're right all this while.
     
  19. Jul 24, 2009 #18

    Mentallic

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    Well, if you think of if as [tex]\frac{1}{0}=\infty[/tex] (it isn't really, but it's close :smile:)

    Then what is the reciprocal? [tex]\frac{1}{\left(\frac{1}{0}\right)}=\frac{1}{\infty}[/tex]
    and in limits, the RHS is assumed to be 0. It is not EXACTLY 0 because it still have a tiny infinitesimal value, but it is small enough to be called 0. This means the LHS is 0 also. You can also see through the division rule: [tex]1 \div \frac{1}{a} = a[/tex] in this case, a=0

    This is, I believe, why an undefined value is not precisely infinite. The reciprocals give zero exactly on the LHS and an infinitesimal on the RHS (which is usually taken as zero in limits).

    It's ok :smile: Better that you expressed your concerns than to either just be ignorant of it or accept the correct answer without question. You actually learn something this way.
     
  20. Jul 24, 2009 #19

    rl.bhat

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    The given problem is tan2x - cot3x = 0
    Then tan2x = cot3x = tan(π/2 -3x)
    So 2x = nπ + ( π/2 - 3x)
    5x = nπ + π/2
    x = (2nπ + π )/10
    Put n = 0, 1, 2, ...until x<2π.
    Ιf you want to solve this problem within five minutes, this is one method.
     
  21. Jul 24, 2009 #20

    jgens

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    Perhaps I'm being pedantic - or misunderstanding what you're saying - but I'm not sure I agree with your interpretation. Multiplicative inverse rules like the one you demonstrated only apply when a ≠ 0. For example, the multiplicative inverse property of R is usually defined such that: For all a in R, if a ≠ 0 then there exists an element a-1 such that a(a-1) = 1. While this may seem arbitrary this restriction is placed (among other reasons) because otherwise R would be inconsistent. Keeping this in mind, the reason that 1/0 ≠ ∞ is because it is not defined, not because working with its multiplicative inverse yields an infinitesimal - which isn't a real number - on one side of the equation and zero on the other.

    Now, to address some comments on limits, it should be noted that 1/0 ≠ ∞ even when working with limits. For example suppose you have Limx -> 0 (1/x). Since x can either be positive or negative the limit diverges to negative infinity or positive infinity. It should also be noted that the notation Limx -> 0 (1/x2) = ∞ is merely shorthand. It is not meant to imply that the limit is actually convergent but rather that the function increases without bound as x -> 0. This is also why statements like 1/∞ = 0 are non-sensical, especially since ∞ is not a real number.

    Perhaps I'm being too pedantic here but hopefully this makes sense. (Hurkyl will probably call me out for not mentioning the projective real numbers).
     
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