# solve equation

• Dec 2nd 2007, 08:51 AM
cumana
solve equation
find the exact solution for

2cosx+1=0
• Dec 2nd 2007, 09:25 AM
Soroban
Hello, cumana!

You must have some idea . . .

Quote:

Find the exact solution for: .$\displaystyle 2\cos x + 1\:=\:0$

We have: .$\displaystyle 2\cos x \:=\:-1\quad\Rightarrow\quad \cos x \:=\:-\frac{1}{2}$

What angle has a cosine of $\displaystyle \frac{1}{2}$ ?
You're expected to know that: .$\displaystyle \cos\frac{\pi}{3} \:=\:\frac{1}{2}$

Since cosine is negative in Quadrants 2 and 3,

. . the answers are: .$\displaystyle x \;=\;\frac{2\pi}{3},\:\frac{4\pi}{3}$

These are the answers in the interval $\displaystyle [0,\,2\pi]$
. . You can generalize them.

• Dec 2nd 2007, 09:28 AM
TKHunny
If you REALLY have no idea how to approach that problem, my inclination is to assume that you have not paid attention in class for the last three years. Can it be?

2cos(x)+1=0

2cos(x) = -1

cos(x) = -1/2

$\displaystyle x\;=\;\frac{2\pi}{3}\;=\;2k\pi,\frac{4\pi}{3}\;+\; 2k\pi\;for\;\pi\;an\; Integer$

I'll just leave it at that. If you don't know what that means or how it happened, I'm going back to my original comments. Let's see where you are. Please have SOME idea.
• Dec 2nd 2007, 09:40 AM
cumana
okie, so i do have some idea, i just didnt know how to go about solving a problem like this one, and my book is rather vague :( thank you very much though