# Solving Trigonometric equations

• Mar 30th 2009, 02:08 AM
mj.alawami
Solving Trigonometric equations
Find all the vlaues of x in $\displaystyle [0,2\pi]$ that are solutions of the equation:

$\displaystyle cos \frac{1}{2}x= \frac{-1}{2}$

Attempt:

$\displaystyle \frac{1}{2}x=arccos \frac{-1}{2}$

$\displaystyle \frac{1}{2}x=120 degrees$
Since cosine is negative in the 2nd and 3rd quadrant
180-120=60
$\displaystyle \frac{1}{2}x=60$
x=120 degrees
No solution

Is my attempt correct?
• Mar 30th 2009, 02:44 AM
Soroban
Hello, mj.alawami!

I understand none of your last paragraph . . .

Quote:

Find all the vlaues of x in $\displaystyle [0,2\pi]$ that are solutions of the equation:

$\displaystyle \cos\tfrac{x}{2}\:=\: \text{-}\tfrac{1}{2}$

Attempt:

$\displaystyle \frac{x}{2}\:=\:\arccos\left(\text{-}\tfrac{1}{2}\right)$

$\displaystyle \frac{x}{2}\:=\:120^o$

Since cosine is negative in the 2nd and 3rd quadrant
180 - 120 = 60
$\displaystyle \tfrac{x}{2}\,=\,60$ . . . . . . . . . . ??
x = 120°
We have: .$\displaystyle \frac{x}{2} \:=\:\arccos\left(\text{-}\tfrac{1}{2}\right)$
Then: .$\displaystyle \frac{x}{2} \;=\;\hdots\: \text{-}240^o,\:\text{-}120^u,\:120^o,\:240^o,\:\hdots$
Hence: .$\displaystyle x \;=\;\hdots\:\text{-}480^o, \:\text{-}240^o,\:240^o,\:480^o,\:\hdots$