Confused about answer wolfram alpha spat out to me.

AI Thread Summary
The Wolfram Alpha output "theta ≈ 0.74559 + 6.28319n for integer n" indicates that the equation has multiple solutions for theta, where n can be any integer. The value 0.74559 is in radians, and converting it to degrees involves multiplying by 180/π. The general solution includes both positive and negative multiples of 2π added to 0.74559. When solving related equations, n can be zero or any integer, reflecting the periodic nature of trigonometric functions. Understanding this helps clarify the infinite solutions to the equation.
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Homework Statement



Ok well I had a real nasty equation in which i could not solve.
I used wolfram alpha to get out this answer "theta ≈ 0.74559 + 6.28319n for integer n".

can someone please tell me what this statement means.

do i assume theta in degrees equals 0.74559(180/pi)?

Thanks for the help!

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Clever_name said:

Homework Statement



Ok well I had a real nasty equation in which i could not solve.
I used wolfram alpha to get out this answer "theta ≈ 0.74559 + 6.28319n for integer n".

can someone please tell me what this statement means.

do i assume theta in degrees equals 0.74559(180/pi)?
That would be one value, with n = 0. The other values are .74559 + 2##\pi##, .74559 + 4##\pi##, .74559 + 6##\pi##, and so on. It would also include negative multiples of 2##\pi##, as in .74559 - 2##\pi##, .74559 - 4##\pi##, .74559 - 6##\pi##, and so on.
 
As to your last question: WolframAlpha gave you answers in radians. Your method of conversion to degrees looks right.
 
So if i was solving an equation say something like x'(theta) =0 would n be zero when solving for theta?
 
Clever_name said:
So if i was solving an equation say something like x'(theta) =0 would n be zero when solving for theta?
n is any integer. You can't say what it "is." It means that the equation has multiple solutions.
 
Ok thanks for clearing that up guys.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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