find the exact solution for
2cosx+1=0
Hello, cumana!
You must have some idea . . .
We have: .$\displaystyle 2\cos x \:=\:-1\quad\Rightarrow\quad \cos x \:=\:-\frac{1}{2}$Find the exact solution for: .$\displaystyle
2\cos x + 1\:=\:0$
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.
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.