Answer
The exact solution of the equation $2\cos 2x+1=0$ on the interval $\left[ 0,2\pi \right)$ is $\frac{\pi }{3}$ , $\frac{5\pi }{3}$ , $\frac{2\pi }{3}$ , or $\frac{4\pi }{3}$.
Work Step by Step
Let us consider the expression given below:
$2\cos 2x+1=0$
Subtract $1$ from both sides:
$\begin{align}
& 2\cos 2x+1-1=0-1 \\
& 2\cos 2x=-1
\end{align}$
And divide by 2 on both sides:
$\begin{align}
& \frac{2\cos 2x}{2}=\frac{-1}{2} \\
& \cos 2x=\frac{-1}{2}
\end{align}$
In the interval $\left[ 0,2\pi \right),$ the cosine function is $\frac{-1}{2}$ at $\frac{2\pi }{3}\text{ and }\frac{4\pi }{3}$.
Therefore,
$\begin{align}
& \cos 2x=\frac{-1}{2} \\
& 2x=\frac{2\pi }{3}
\end{align}$
Or,
$\begin{align}
& \cos 2x=\frac{-1}{2} \\
& 2x=\frac{4\pi }{3}
\end{align}$
Since the period of the cosine function is $2\pi $, the general solution of the equation is:
$\begin{align}
& 2x=\frac{2\pi }{3}+2n\pi \\
& x=\frac{\pi }{3}+n\pi
\end{align}$
Or,
$\begin{align}
& 2x=\frac{4\pi }{3}+2n\pi \\
& x=\frac{2\pi }{3}+n\pi
\end{align}$
To get different solutions, put $\text{ }n=0,1,2,3\ldots $
When $n=0$ ,
$x=\frac{\pi }{3}\text{ or }x=\frac{2\pi }{3}$
When $n=1$
$\begin{align}
& x=\frac{\pi }{3}+\pi \\
& =\frac{4\pi }{3}
\end{align}$
Or,
$\begin{align}
& x=\frac{2\pi }{3}+\pi \\
& =\frac{2\pi +3\pi }{3} \\
& =\frac{5\pi }{3}
\end{align}$
Put the other values of $n$; then the solution becomes outside the interval $\left[ 0,2\pi \right).$
Thus, the exact solution of the equation $2\cos 2x+1=0$ on the interval $\left[ 0,2\pi \right)$ is $\frac{\pi }{3}$ , $\frac{5\pi }{3}$ , $\frac{2\pi }{3}$ , or $\frac{4\pi }{3}$.
