# Thread: Finding numbers which satisfy trigonometric equations

1. ## Finding numbers which satisfy trigonometric equations

This is the question:

Find the numbers(s) x in the interval [0, 2Pi] which satisfies the equation,
tan x / 2 = 1

I multiplied both sides by 2, and solved for x using my calculator:
tan x = 2
x = 63.43

However, the answer is Pi/2. What have I done wrong? 63.43 degrees is a long shot from 90 degrees. :/

2. Originally Posted by Glitch
This is the question:

Find the numbers(s) x in the interval [0, 2Pi] which satisfies the equation,
tan x / 2 = 1

I multiplied both sides by 2, and solved for x using my calculator:
tan x = 2
x = 63.43

However, the answer is Pi/2. What have I done wrong? 63.43 degrees is a long shot from 90 degrees. :/
$tan\frac{x}{2} \neq \frac{1}{2}tan{x}$ (edit: it equals only at $x=2\pi n$

Anyway....

Hint:

When $sinx$ and $cosx$ get the same values, and way solving my hint answer your question?

3. Originally Posted by Glitch
This is the question:

Find the numbers(s) x in the interval [0, 2Pi] which satisfies the equation,
tan x / 2 = 1

I multiplied both sides by 2, and solved for x using my calculator:
tan x = 2
x = 63.43

However, the answer is Pi/2. What have I done wrong? 63.43 degrees is a long shot from 90 degrees. :/
Are you sure it's not

$\tan{\left(\frac{x}{2}\right)} = 1$?

If so, you should get

$\frac{x}{2} = \tan^{-1}(1)$

$\frac{x}{2} = \frac{\pi}{4} + \pi n$ where $n \in \mathbf{Z}$

$x = \frac{\pi}{2} + 2\pi n$.

4. Originally Posted by Prove It
Are you sure it's not

$\tan{\left(\frac{x}{2}\right)} = 1$?
Ahh, that may be it. The textbook wrote it exactly how I did in my first post (which isn't very clear). Thank you.