Help Notes: 2cosx + 1 = 0 find x values for the interval [0, 2pi]

Thursday, 15 July 2010

We
have to find the value of x for $\displaystyle{2}{\cos{{x}}}+{1}={0}$ in the interval $[0,2\pi]$

So,

$\displaystyle{2}{\cos{{x}}}+{1}={0}$

In other words,
$\displaystyle{2}{\cos{{x}}}=-{1}$

$cosx=\frac{-1}{2}$

The solution will be of the
form:

$x=\pm cos^{-1}(\frac{-1}{2})+2\pi n$ where n=0,1,2,. . .

For example, $x=\pm \frac{2\pi}{3} +2\pi n , \ n=0,1,2,...$

When
,

$n=0 : x=\pm \frac{2\pi}{3}$

$n=1:\ x=$
\frac{2\pi}{3}+2\pi = \frac{8\pi}{3}

or ,
$x=\frac{-2\pi}{3}+2\pi=\frac{4\pi}{3}$

However, we have to find the angle in
the interval $[0,2\pi]$

As a result, the answer is:

$x=\frac{2\pi}{3} \ and \ x=\frac{4\pi}{3}$

Help Notes