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Inverse trigonometric equation 
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#1
Jan513, 10:07 AM

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1. The problem statement, all variables and given/known data
Solve [itex]cot^{1} x  cot^{1} (n^2  x+1)=cot^1(n1)[/itex] 2. Relevant equations 3. The attempt at a solution I can write the above equation as θ+α=β where the symbols represent the respective inverse functions. Now I take tan of both sides. Simplifying I get [itex](n^2+12x)(n1) = x(n^2x+1)[/itex] 


#2
Jan513, 01:25 PM

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Are you asking how to solve that equation?



#3
Jan513, 01:43 PM

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[itex](n^2+12x)(n1) = 1+x(n^2x+1)[/itex] 


#4
Jan613, 12:53 AM

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Inverse trigonometric equation



#5
Jan613, 12:53 AM

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#6
Jan613, 01:36 AM

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Get everything to the left side. You have a quadratic equation in x. 


#7
Jan613, 08:54 AM

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#8
Jan613, 02:43 PM

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What do you get for the discriminant ? 


#9
Jan713, 02:32 AM

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D=[itex]n^4+6n^28n+2[/itex] 


#10
Jan713, 08:42 AM

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I got it because, [itex]\displaystyle \left(1+\frac{1}{x(n^2x+1)}\right)x(n^2x+1)=x(n^2x+1)+1\ .[/itex] How did you not get it? 


#11
Jan813, 06:23 AM

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[itex] x= \dfrac{n(n+1)}{2} [/itex] And yes you said right. There was a calculation error in my solution. It should have that extra 1 


#12
Jan813, 01:53 PM

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#13
Jan913, 01:41 AM

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I still get a quadratic equation in x, so that can't be the solution.
[itex]\displaystyle \cot(A+B)=\frac{\cot(A)\,\cot(B)1}{\cot(A)+\cot(B)}[/itex] 


#14
Jan913, 06:34 AM

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[itex] 4n^3+6n^28n+9[/itex] 


#15
Jan913, 12:22 PM

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[itex]n^4+6n^28n+9\ .[/itex] I just rechecked it using WolframAlpha on [itex](n^2+12x)(n1) = 1+x(n^2x+1)[/itex] and it confirmed my result. 


#16
Jan913, 12:51 PM

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#17
Jan913, 02:53 PM

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#18
Jan1013, 10:05 AM

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