1. The problem is to determine which of the points A, B, C, or D lie on the graph of the function $f(x) = \tan(5x)$. 2. To check if a point $(x,y)$ lies on the graph, we substitute $x$ into the function and see if $f(x) = y$. 3. Recall that $\tan(\theta)$ is the tangent function, and we will use exact values for tangent at special angles. 4. Check point A: $x = -\frac{\pi}{6}$, $y = \frac{2}{3}\sqrt{3}$. Calculate $f\left(-\frac{\pi}{6}\right) = \tan\left(5 \times -\frac{\pi}{6}\right) = \tan\left(-\frac{5\pi}{6}\right)$. Since $\tan(-\theta) = -\tan(\theta)$, $$\tan\left(-\frac{5\pi}{6}\right) = -\tan\left(\frac{5\pi}{6}\right).$$ $\tan\left(\frac{5\pi}{6}\right) = \tan\left(\pi - \frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$. Therefore, $$f\left(-\frac{\pi}{6}\right) = -\left(-\frac{1}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}}.$$ Compare with $y = \frac{2}{3}\sqrt{3}$. Since $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ and $\frac{2}{3}\sqrt{3} = \frac{2\sqrt{3}}{3}$, these are not equal. So point A is not on the graph. 5. Check point B: $x = -\frac{\pi}{3}$, $y = \frac{1}{2}\sqrt{3}$. Calculate $f\left(-\frac{\pi}{3}\right) = \tan\left(5 \times -\frac{\pi}{3}\right) = \tan\left(-\frac{5\pi}{3}\right)$. Since $\tan$ has period $\pi$, $$\tan\left(-\frac{5\pi}{3}\right) = \tan\left(-\frac{5\pi}{3} + 2\pi\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}.$$ Compare with $y = \frac{1}{2}\sqrt{3}$. They are not equal, so point B is not on the graph. 6. Check point C: $x = \frac{2\pi}{3}$, $y = \sqrt{3}$. Calculate $f\left(\frac{2\pi}{3}\right) = \tan\left(5 \times \frac{2\pi}{3}\right) = \tan\left(\frac{10\pi}{3}\right)$. Reduce angle modulo $\pi$: $$\frac{10\pi}{3} - 3\pi = \frac{10\pi}{3} - \frac{9\pi}{3} = \frac{\pi}{3}.$$ So, $$f\left(\frac{2\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}.$$ This matches $y = \sqrt{3}$. So point C is on the graph. 7. Check point D: $x = \frac{\pi}{6}$, $y = -\frac{1}{3}\sqrt{3}$. Calculate $f\left(\frac{\pi}{6}\right) = \tan\left(5 \times \frac{\pi}{6}\right) = \tan\left(\frac{5\pi}{6}\right)$. As before, $$\tan\left(\frac{5\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.$$ Compare with $y = -\frac{1}{3}\sqrt{3} = -\frac{\sqrt{3}}{3}$. They are equal, so point D is on the graph. Final answer: Points C and D lie on the graph of $f(x) = \tan(5x)$.