Thread: Solving for X - Where am I off?

1. Solving for X - Where am I off?

If I have $tan{x}+1=0$, then I have $tan{x} = -1$

well, tan is equal to 1 in $\frac{\pi}{4}$. It's equal to -1 in quadrant 2 and 4.

So $x = \frac{3}{4}{\pi}+2n{\pi}$ where n is all reals.

but why isn't it also $x = \frac{7}{4}{\pi}+2n{\pi}$, since tan is negative in the 4th quadrant as well?

2. Re: Solving for X - Where am I off?

Actually it's $\displaystyle \tan{x} = 1$, and the tangent function is positive in the first and third quadrants.

$\displaystyle x = \frac{\pi}{4} + \pi n$ where $\displaystyle n \in \mathbf{Z}$.

3. Re: Solving for X - Where am I off?

Sorry it my quickness to get the latex correct I wrote the problem out wrong. its supposed to have been TanX+1, not tanx-1

4. Re: Solving for X - Where am I off?

Well to answer your question, both answers are correct. They can be written in a more compact way as $\displaystyle x = \frac{3\pi}{4} + \pi n$ where $\displaystyle n \in \mathbf{Z}$. The period of the tangent function is $\displaystyle \pi$, not $\displaystyle 2\pi$.

5. Re: Solving for X - Where am I off?

Originally Posted by DamenFaltor
Sorry it my quickness to get the latex correct I wrote the problem out wrong. its supposed to have been TanX+1, not tanx-1
I suppose you still have the same difficulty as before:

The tangent function is periodic with a period of π not 2π.

6. Re: Solving for X - Where am I off?

Ah yes, I forgot about the periodicity of Tan being different.

ok so to provide:

${x}= \frac{3}{4}{\pi} + {n}{\pi} ; \frac{7}{4}{\pi} + {n}{\pi}$

Would be a correct answer?

7. Re: Solving for X - Where am I off?

The second of those is redundant, since $\displaystyle \frac{3\pi}{4} + \pi = \frac{7\pi}{4}$.