Solving Trigonometric equations

**mj.alawami** · March 30th 2009, 02:08 AM

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

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

Attempt:

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

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

Is my attempt correct?

**Soroban** · March 30th 2009, 02:44 AM

Hello, mj.alawami!

I understand none of your last paragraph . . .

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

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

Attempt:

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

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

Since cosine is negative in the 2nd and 3rd quadrant
2nd quadrant
180 - 120 = 60
$\tfrac{x}{2}\,=\,60$ . . . . . . . . . . ??
x = 120°
3rd quadrant
No solution

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

Then: . $\frac{x}{2} \;=\;\hdots\: \text{-}240^o,\:\text{-}120^u,\:120^o,\:240^o,\:\hdots$

Hence: . $x \;=\;\hdots\:\text{-}480^o, \:\text{-}240^o,\:240^o,\:480^o,\:\hdots$

Therefore: no solutions.

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