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Find real solutions to an equation

  1. Nov 23, 2015 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    log(cosx)sinx = 4*log(sinx)cosx
    2. Relevant equations
    3. The attempt at a solution
    i tried to solve it and uploaded my work but my last part reads
    (cosx)^2 = sinx
    (cosx)^(-2) = sinx
    its weird that i dont see where i have made a mistake
     
  2. jcsd
  3. Nov 23, 2015 #2

    SammyS

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    First of all:
    Do you really mean (log(cos x))⋅sin x = 4⋅(log(sin x))⋅cos x ? (Mathematically speaking, that is what you wrote.)

    Furthermore, it's difficult to say where you went wrong, without you giving more detail in your workings.
     
  4. Nov 23, 2015 #3

    diredragon

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    What is in the brackets i meant to be the base.
     
  5. Nov 23, 2015 #4

    SammyS

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    So, your initial problem must have had:
    log(cosx)(sinx) = 4*log(sinx)(cosx) .​
     
  6. Nov 23, 2015 #5

    diredragon

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    yes and im trying to show my work but the pic wont upload. Ill try to fix it
     
  7. Nov 23, 2015 #6

    diredragon

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  8. Nov 23, 2015 #7

    SammyS

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    Sideways image.

    I get the same result using change of base formula.

    Rewrite that second result as:
    1 = (sin x)(cos2 x)

    I think only the first one has real number solutions.
     
  9. Nov 23, 2015 #8

    Mark44

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    As already noted, the above appears to be ##\log(\cos x) \cdot \sin x = 4 \log(\sin x) \cdot \cos x##
    As it turns out, you intended the log bases to be cos(x) and sin(x), respectively

    To make you work understandable, use either BBCode subscripts or LaTeX to write these expressions. Below are the a couple of examples of each, showing how they are formed and how they appear.
    BBCode: log[SUB]cos(x)[/SUB](sin(x)) - renders as logcos(x)(sin(x) -- see https://www.physicsforums.com/help/bb-codes
    LaTeX: ##\log_{\cos(x)}\sin(x)## - renders as ##\log_{\cos(x)}\sin(x)## -- see https://www.physicsforums.com/help/latexhelp/
     
  10. Nov 23, 2015 #9

    diredragon

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    I get
    -(sinx)^2 + 1 = sinx
    And
    1 = sinx - (sinx)^3 for the two cases
    How do i now find the real solutions out of this?
     
  11. Nov 23, 2015 #10

    Samy_A

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    The first is a quadratic equation (in sin(x)).
    The second one you better leave in the form @SammyS used, and try to see why he thought that it doesn't have a real solution.
     
  12. Nov 24, 2015 #11

    diredragon

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    From the first quadratic equation i get sinx=(-1+-(5)^(1/2))/2 the only real solution is + one because log is not defined for negative numbers. Using the identity i get cosx to be the same. How now do i obtain the real solutions? Is this correct?
     
  13. Nov 24, 2015 #12

    Samy_A

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    I also get ##\sin(x)=\frac {\sqrt{5}-1}{2}## as solution for the quadratic equation.You can indeed reject the other solution to the quadratic equation, ##\frac {-\sqrt{5}-1}{2}##.

    I don't understand your remark about cos(x).

    Anyway, ##\sin(x)=\frac {\sqrt{5}-1}{2}## should give you the real solutions (I think the solutions will have to include the ##\arcsin## function, but I'm not sure, maybe someone can find a nice solution without ##\arcsin##).
     
    Last edited: Nov 24, 2015
  14. Nov 24, 2015 #13

    diredragon

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    I made a mistake concerning cosin i figured cosx = (1 - sinx)^1/2 but that doesnt help me. The solutions of this problem somehow involves specific angles.
    I have the solution but not the process of solving. It reads (π/6, π/4].
    Any chance someone helping us with this?
     
  15. Nov 24, 2015 #14

    Samy_A

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    How can π/4 be a solution?
    cos(π/4)=sin(π/4)=##\sqrt2/2##. Let's set ##y=\sqrt2/2##, then your original equation becomes ##\log_yy=4*log_yy##, or ##1=4##. :wink:

    π/6 doesn't look like a solution either (did a quick check in Excel).
     
  16. Nov 24, 2015 #15

    diredragon

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    I dont understand either. It says in the problem to find all real solutions to that equation and it gives its solution as
    All real solutions are in the range from pi/6 to pi/4 discluding the pi/6 and including the pi/4. I have no idea how they got to that.
     
  17. Nov 24, 2015 #16

    Samy_A

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    Ow, you meant that the solution lies in that interval. Yes, that is the case, up to the 2πn terms.
     
    Last edited: Nov 24, 2015
  18. Nov 24, 2015 #17

    diredragon

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    How did it get to that? How to know which interval includes the real solutions to this?
     
  19. Nov 24, 2015 #18

    Samy_A

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    We know that ##\sin(x)=\frac {\sqrt{5}-1}{2}=0.618 ...##
    ##\arcsin(0.618)=0.666 ... ## in radians, and that is approximately 38°. (I take the solution in the first quadrant, as sin(x) and cos(x) must both be positive.)
    So the solution indeed lies between π/6 (30°) and π/4 (45°).

    Of course, once you have a solution x, x+2πn (where n is an integer) will also be a solution.
     
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