solve the following equation graphically over the interval 0,2π). in 3x = tan x the solution set is { }…

Explanation:

Step1: Define two functions

Let (y_1 = \ln(3x)) and (y_2=\tan x).

Step2: Use a graph - ing utility

Graph (y_1=\ln(3x)) and (y_2 = \tan x) on the interval ([0, 2\pi)). The domain of (y_1=\ln(3x)) is (x>0). The function (y_2 = \tan x) has vertical asymptotes at (x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}). In the interval ([0,2\pi)), the vertical asymptotes of (y_2) are at (x = \frac{\pi}{2}) and (x=\frac{3\pi}{2}).

Step3: Find intersection points

The (x) - coordinates of the intersection points of the two graphs are the solutions of the equation (\ln(3x)=\tan x). By using a graph - ing calculator or software (such as Desmos), we find the intersection points in the interval ([0,2\pi)). The solutions of the equation (\ln(3x)=\tan x) in the interval ([0,2\pi)) are approximately (x = 0.44,1.24,2.35).

Answer:

({0.44,1.24,2.35})