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I'm having a little trouble determining the solutions of:

$$\tan x - 5 = 0,~ \text{at the interval}~ x \in (0, Pi)$$

I figured that $tanx = 5$ and when you draw the graph you have to draw it between $0$ and $2 \pi$ on the $x$ axis and below $5$ on the $y$ axis. However I'm not really sure how to move on from there.

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up vote 1 down vote accepted

The problem isn't asking you to graph this, but to solve it.

You could solve it graphically by plotting $y = tan(x) - 5$, and then seeing where (for what values of $x$) it crosses the $x$-axis. (Because the $x$-axis is where $y=0$)

Like this

You could also solve it algebraically, by first adding $5$ to both sides, $$ \tan(x) = 5 $$ and then solving using arctangent: $$ x = \arctan(5) + k\pi = 1.37340077 + k\pi $$ where $k$ is any integer. You'll find that the values of $x$ given for $k=0$ and $k=1$ are between $0$ and $2\pi$.

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sorry i havent learned arctangent and i was looking through the unit and it doesnt mention it – Exikle Mar 13 '13 at 4:51
and i accidently wrote 2Pi, its actually just Pi – Exikle Mar 13 '13 at 4:55
i have edited the origianl post to make it pi – Exikle Mar 13 '13 at 4:58
It might refer to it as $\tan^{-1}$, but otherwise I'm guessing you are intended to solve it graphically. Do you do a lot of solution by graph in your class? If you have a TI-83/84, plot $y1 = \tan(x) - 5$, change your window to xmin=$0$ xmax=$2\pi$, and then use the zero tool to find where it crosses the $x$-axis. – BaronVT Mar 13 '13 at 5:00

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