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anonymous

  • one year ago

solve cotX^2+4cotX-7=0

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  1. anonymous
    • one year ago
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    find the general solution

  2. anonymous
    • one year ago
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    @Michele_Laino

  3. SolomonZelman
    • one year ago
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    you can at first say, let \(\cot(x)=a\), which will give you the following quadratic equation, \(a^2+4a-7=0\)

  4. SolomonZelman
    • one year ago
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    Solve this equation for a, and after you find the solutions for a, you can find the value of x, using the fact that \(a=\cot(x)\)

  5. SolomonZelman
    • one year ago
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    (well, not the value of x, but a set of solutions, but I hope you see what I mean)

  6. anonymous
    • one year ago
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    what about the general solution?

  7. mathstudent55
    • one year ago
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    When you write \(\cot x^2\), do you really mean \(\cot^2 x\) ?

  8. anonymous
    • one year ago
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    the latter

  9. SolomonZelman
    • one year ago
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    yes, mathstudent55 I would assume so:) I didn't even notice that mistake (if it is a mistake)

  10. SolomonZelman
    • one year ago
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    have you solved for "a" (using a substitution of a=cot(x) that I offered) ?

  11. anonymous
    • one year ago
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    yes but that just gave me -2 +or-sqrt(44)

  12. SolomonZelman
    • one year ago
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    i got mixed up for some reason

  13. SolomonZelman
    • one year ago
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    a²-4a-7=0 a²-4a=7 a²-4a+4=7+4 a²-4a+4=11 (a-2)²=11 a-2=±√11 a=-2±√11 that is what I get

  14. anonymous
    • one year ago
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    its +4a

  15. SolomonZelman
    • one year ago
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    oh, tnx for catching me

  16. anonymous
    • one year ago
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    (y)

  17. SolomonZelman
    • one year ago
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    oh, my fault, i did the step incorrectly

  18. SolomonZelman
    • one year ago
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    a²-4a-7=0 a²-4a=7 a²-4a+4=7+4 a²-4a+4=11 (a-2)²=11 a-2=±√11 a=2±√11 that is what I get if -4a ------------------ thus with +4a i would get now with +4a i get a²+4a-7=0 a²+4a=7 a²+4a+4=7+4 a²+4a+4=11 (a+2)²=11 a+2=±√11 a=-2±√11

  19. SolomonZelman
    • one year ago
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    and by the quadratic formula a²+4a-7=0 \(\large\color{black}{ \displaystyle \frac{-4\pm\sqrt{(4)^2-4(-7)(1)} }{2} }\) \(\large\color{black}{ \displaystyle \frac{-4\pm\sqrt{44} }{2} }\) \(\large\color{black}{ \displaystyle \frac{-4}{2}\pm\frac{\sqrt{44} }{2} }\)

  20. SolomonZelman
    • one year ago
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    see I am getting the same thing a = -2±√11

  21. anonymous
    • one year ago
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    okay, yay thanks, i get it now!

  22. SolomonZelman
    • one year ago
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    yes, so \(\cot(x)=-2+\sqrt{11}\) or \(\cot(x)=-2-\sqrt{11}\)

  23. SolomonZelman
    • one year ago
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    \(\cot^{-1}(-2+\sqrt{11})=a\) or \(\cot^{-1}(-2-\sqrt{11})=a\)

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