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anonymous
 one year ago
solve cotX^2+4cotX7=0
anonymous
 one year ago
solve cotX^2+4cotX7=0

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0find the general solution

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3you can at first say, let \(\cot(x)=a\), which will give you the following quadratic equation, \(a^2+4a7=0\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3Solve this equation for a, and after you find the solutions for a, you can find the value of x, using the fact that \(a=\cot(x)\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3(well, not the value of x, but a set of solutions, but I hope you see what I mean)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about the general solution?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0When you write \(\cot x^2\), do you really mean \(\cot^2 x\) ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3yes, mathstudent55 I would assume so:) I didn't even notice that mistake (if it is a mistake)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3have you solved for "a" (using a substitution of a=cot(x) that I offered) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but that just gave me 2 +orsqrt(44)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3i got mixed up for some reason

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3a²4a7=0 a²4a=7 a²4a+4=7+4 a²4a+4=11 (a2)²=11 a2=±√11 a=2±√11 that is what I get

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3oh, tnx for catching me

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3oh, my fault, i did the step incorrectly

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3a²4a7=0 a²4a=7 a²4a+4=7+4 a²4a+4=11 (a2)²=11 a2=±√11 a=2±√11 that is what I get if 4a  thus with +4a i would get now with +4a i get a²+4a7=0 a²+4a=7 a²+4a+4=7+4 a²+4a+4=11 (a+2)²=11 a+2=±√11 a=2±√11

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3and by the quadratic formula a²+4a7=0 \(\large\color{black}{ \displaystyle \frac{4\pm\sqrt{(4)^24(7)(1)} }{2} }\) \(\large\color{black}{ \displaystyle \frac{4\pm\sqrt{44} }{2} }\) \(\large\color{black}{ \displaystyle \frac{4}{2}\pm\frac{\sqrt{44} }{2} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3see I am getting the same thing a = 2±√11

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, yay thanks, i get it now!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3yes, so \(\cot(x)=2+\sqrt{11}\) or \(\cot(x)=2\sqrt{11}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3\(\cot^{1}(2+\sqrt{11})=a\) or \(\cot^{1}(2\sqrt{11})=a\)
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