Question

What are the zeros of$f\left(x\right)=2\cos \left(3x\right)+1$ on the interval $[0,2\mathrm{\pi }]$?

Short answer

The zeros of the given equation is$\left\{\frac{2\mathrm{\pi }}{9},\frac{4\mathrm{\pi }}{9}\right\}$

Step-by-step solution

Step 1. Given Information 

We have to find the zeros of$f\left(x\right)=2\cos \left(3x\right)+1$ on the interval $[0,2\mathrm{\pi }]$.

Step 2. The zeros are the x values where the graph intersects the x-axis. 

To find the zeros, replace $f\left(x\right)$with $0$and solve for $x$.

Let $3x=y$

Now the equation is

$$\begin{gathered} 0=2\cos y+1 \\ 2\cos y+1=0 \end{gathered}$$

Step 3. Subtract 1 on both side 

$$\begin{gathered} 2\cos y+1-1=0-1 \\ 2\cos y=-1 \end{gathered}$$

Divide by 2 on both side

$$\begin{gathered} \frac{2}{2}\cos y=-\frac{1}{2} \\ \cos y=-\frac{1}{2} \end{gathered}$$

Step 4. Solving each equation in the interval [0,2π], we obtain 

$y=\frac{2\mathrm{\pi }}{3}\mathrm{or}\mathrm{y}=\frac{4\mathrm{\pi }}{3}$

Since $y=3x$,So

$$\begin{gathered} 3x=\frac{2\mathrm{\pi }}{3}\mathrm{or}3\mathrm{x}=\frac{4\mathrm{\pi }}{3} \\ \frac{3}{3}x=\frac{2\mathrm{\pi }}{3}·\frac{1}{3}\mathrm{or}\frac{3}{3}\mathrm{x}=\frac{4\mathrm{\pi }}{3}·\frac{1}{3} \\ x=\frac{2\mathrm{\pi }}{9}\mathrm{or}\mathrm{x}=\frac{4\mathrm{\pi }}{9} \end{gathered}$$

So the zeros are$\left\{\frac{2\mathrm{\pi }}{9},\frac{4\mathrm{\pi }}{9}\right\}$