## Precalculus (6th Edition) Blitzer

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 $\frac{4\pi }{3}$.
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 $\frac{4\pi }{3}$.