question 3 of 10 how many solutions does the equation cos(6x) = 1/2 have on the interval (0, 2π?

Explanation:

Step1: Recall cosine - inverse relation

If $\cos\theta=\frac{1}{2}$, then $\theta = 2k\pi\pm\frac{\pi}{3}$, where $k\in\mathbb{Z}$. So for $\cos(6x)=\frac{1}{2}$, we have $6x = 2k\pi\pm\frac{\pi}{3}$.

Step2: Solve for $x$

$x=\frac{2k\pi\pm\frac{\pi}{3}}{6}=\frac{k\pi}{3}\pm\frac{\pi}{18}$.

Step3: Find valid $k$ values for the interval $(0,2\pi]$

For the positive part $x=\frac{k\pi}{3}+\frac{\pi}{18}$: When $k = 0$, $x=\frac{\pi}{18}\in(0,2\pi]$. When $k = 1$, $x=\frac{\pi}{3}+\frac{\pi}{18}=\frac{6\pi + \pi}{18}=\frac{7\pi}{18}\in(0,2\pi]$. When $k = 2$, $x=\frac{2\pi}{3}+\frac{\pi}{18}=\frac{12\pi+\pi}{18}=\frac{13\pi}{18}\in(0,2\pi]$. When $k = 3$, $x=\pi+\frac{\pi}{18}=\frac{18\pi+\pi}{18}=\frac{19\pi}{18}\in(0,2\pi]$. When $k = 4$, $x=\frac{4\pi}{3}+\frac{\pi}{18}=\frac{24\pi+\pi}{18}=\frac{25\pi}{18}\in(0,2\pi]$. When $k = 5$, $x=\frac{5\pi}{3}+\frac{\pi}{18}=\frac{30\pi+\pi}{18}=\frac{31\pi}{18}\in(0,2\pi]$. For the negative part $x=\frac{k\pi}{3}-\frac{\pi}{18}$: When $k = 1$, $x=\frac{\pi}{3}-\frac{\pi}{18}=\frac{6\pi - \pi}{18}=\frac{5\pi}{18}\in(0,2\pi]$. When $k = 2$, $x=\frac{2\pi}{3}-\frac{\pi}{18}=\frac{12\pi-\pi}{18}=\frac{11\pi}{18}\in(0,2\pi]$. When $k = 3$, $x=\pi-\frac{\pi}{18}=\frac{18\pi - \pi}{18}=\frac{17\pi}{18}\in(0,2\pi]$. When $k = 4$, $x=\frac{4\pi}{3}-\frac{\pi}{18}=\frac{24\pi-\pi}{18}=\frac{23\pi}{18}\in(0,2\pi]$. When $k = 5$, $x=\frac{5\pi}{3}-\frac{\pi}{18}=\frac{30\pi-\pi}{18}=\frac{29\pi}{18}\in(0,2\pi]$. When $k = 6$, $x = 2\pi-\frac{\pi}{18}=\frac{36\pi-\pi}{18}=\frac{35\pi}{18}\in(0,2\pi]$. Counting all these solutions, we find there are 12 solutions.

Answer:

12