chapter 9: review score: 120/280 answered: 12/28 question 22 0/10 pts 5 99 details find all solutions of the…

Explanation:

Step1: Isolate cosine function

Divide both sides of $2\cos3x = 1$ by 2. We get $\cos3x=\frac{1}{2}$.

Step2: Find general solutions

The general solution of $\cos\theta=\frac{1}{2}$ is $\theta = 2k\pi\pm\frac{\pi}{3}$, where $k\in\mathbb{Z}$. So for $\cos3x=\frac{1}{2}$, we have $3x = 2k\pi\pm\frac{\pi}{3}$, then $x=\frac{2k\pi}{3}\pm\frac{\pi}{9}$.

Step3: Find solutions in $[0,\pi)$

When $k = 0$, $x_1=\frac{\pi}{9}$ (using the plus - sign in $x=\frac{2k\pi}{3}\pm\frac{\pi}{9}$). When $k = 1$, $x_2=\frac{2\pi}{3}-\frac{\pi}{9}=\frac{6\pi - \pi}{9}=\frac{5\pi}{9}$. When $k = 1$, using the plus - sign, $x_3=\frac{2\pi}{3}+\frac{\pi}{9}=\frac{6\pi+\pi}{9}=\frac{7\pi}{9}$.

Answer:

$x_1=\frac{\pi}{9}$, $x_2=\frac{5\pi}{9}$, $x_3=\frac{7\pi}{9}$