Question 1183391
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Your statement "The answer is *[tex \Large Ak\pi] where k is any integer." is a false statement.  That is only one of three solutions to the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan^5(x)\ -\ 9\tan(x)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(x)\(\tan^4(x)\ -\ 9\)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(x)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 0\ +\ k\pi\ :\ k\ \in\ \mathbb{Z}]


meaning that *[tex \Large A\ =\ 1] in the context of your given answer.


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan^4(x)\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(x)\ =\ \pm\sqrt{3}]


If *[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(x)\ =\ \sqrt{3}]


Then *[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\pi}{3}\ +\ k\pi\ :\ k\ \in\ \mathbb{Z}]


If *[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(x)\ =\ -\sqrt{3}]


Then *[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{2\pi}{3}\ +\ k\pi\ :\ k\ \in\ \mathbb{Z}]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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