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Solve trig multiply equation

  1. Mar 1, 2015 #1
    Anyone know a way to calculate x=tan(x)cosh2(x) ? I think I should know - but just blank at the moment.
  2. jcsd
  3. Mar 1, 2015 #2
    If you want the numbers for which the equation holds, I think you may want (and have) to do it by plotting, or using some numeric method.

    It has infinite solutions, the first one has be zero since you have [itex]\tan(0)=0[/itex], some solutions are
    x = 3.36922, 6.3062, 9.4263, 12.5665, 15.708
  4. Mar 1, 2015 #3
    I suspected that, did you use a graphing tool of some sort to fund these values? (If so, what ....). Thanks
  5. Mar 1, 2015 #4
    Terribly sorry, I had a typo - hyperbolic tan, it should be x=tanh(x)cosh2(x) [or x = 1/2 cosh(2x) ]
  6. Mar 2, 2015 #5
    [itex]\tanh(x)[/itex] goes from zero to one, [itex]\cosh^2(x)[/itex] blows up to infinity, the only solution is x=0.

    Also, [itex]\tanh(x)\cosh^2(x) = \frac{1}{2}\sinh(x) [/itex] and [itex] x = \frac{1}{2}\cosh(x) [/itex] has no solution
  7. Mar 2, 2015 #6
    Sorry, yes, I want to solve x= ½sinh(x). IE 2x = x + x3/3! + x5/5!. Dividing by x gives 2 = 1 + x2/6 + v4/120 + .....
    Letting u = x2, I have a quadratic of u2 + 20u - 120 = 0 and the positive root of this is 4.8324, giving x=2.2098 ?
  8. Mar 2, 2015 #7


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    Not quite. When you truncated the infinite series, you lost some accuracy.

    BTW, you can check these calculations yourself, either with a calculator or online, at say Wolfram Alpha.
  9. Mar 2, 2015 #8
    As already said, thats an approximation since you are not using the whole series. A more precise result using Mathematica is 2.1773189849653067526.
    By the way, I made a typo, I meant [itex]\frac{1}{2}\sinh(2x)[/itex] wich solution is x=0
  10. Mar 2, 2015 #9
    I confess I dropped the x7 and subsequent terms because I dont know how to solve the more complex equation that would result :-( So if I had (say) x8/9! + x6/7! + x4/5! + x3/3! - 1 =0, how would I approach solving that equation without using something like mathematica?
  11. Mar 2, 2015 #10


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    In general, there is no formula for solving polynomial equations of degree 5 and greater. Unless you stumble on some fortuitous factorization, you're basically left with numerical techniques, like iteration or applying Newton's method. Applying things like the Rational Root Theorem and Descartes Rule of Signs may indicate how many real roots exist and which trial values to try, but it's basically trial and error.
  12. Mar 4, 2015 #11
    Thanks, thats all been very clear and helpful.
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