# Finding all solutions to $\tan^5x - 9\tan{x} = 0$

I am stuck when it comes to finding the end value of a trig function. I have the following question:

$$\tan^5x - 9\tan{x} = 0$$

I worked the problem and got:

$$\tan x = 0\\ \tan^4x-9 = 0\\ x = 0, \pi, \frac {\pi}{3}, \frac {2\pi}{3}, \frac {4\pi}{3}, \frac {5\pi}{3}$$

My book answer is $x = \frac {\pi k}{3}$ how do you get that? I understand that tan uses $\pi$ and sin, cos use $2\pi$ but I'm not sure how they got to that answer.

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Could you also state the domain of $k$? Eg. $k = 2, 4, 6, ...$. – bryansis2010 Feb 13 '13 at 5:01
Look at the answers you got. Are they not all of the form $\pi k/3$, for various values of $k$? – Gerry Myerson Feb 13 '13 at 5:04

$$\tan^4x=9\Longrightarrow \tan^2x= 3\Longrightarrow \tan x=\pm\sqrt 3=\pm\frac{\frac{\sqrt 3}{2}}{\frac{1}{2}}=\pm\frac{\sin\frac{\pi}{3}}{\cos\frac{\pi}{3}}$$

Can you see it now?

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$\tan^2x=3\implies \cos2x=\frac{1-\tan^2x}{1+\tan^2x}=\frac{1-3}{1+3}=-\frac12=\cos\frac{2\pi}3$

$2x=2n\pi\pm \frac{2\pi}3=\frac{2\pi}3(3n\pm1)$ where $n$ is any integer.

$x=\frac{\pi}3(3n\pm1)$

Now, we need $0\le \frac{\pi}3(3n+1)<2\pi\implies 0\le3n+1<6\implies n=0,1$

For $n=0, x=\frac{\pi}3(3n+1)= \frac{\pi}3; n=1\implies x=\frac{\pi}3(3n+1)= \frac{4\pi}3$

Similarly, we need $0\le \frac{\pi}3(3n-1)<2\pi\implies 0\le3n-1<6\implies n=1,2$

For $n=1, x=\frac{\pi}3(3n-1)= \frac{2\pi}3; n=2\implies x=\frac{\pi}3(3n-1)= \frac{5\pi}3$

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You made an error in your work. e.g. when you solved $\tan x = 0$, you only found the two solutions $0$ and $\pi$. But you missed all of the other solutions like $2 \pi$, $3 \pi$, and $-1872 \pi$. The set of all solutions to $\tan x = 0$ is the set of all values $\pi v$ where $v$ ranges over all integers.

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