The problem is to find the solution set of the inequality $(6x + 1)(4x - 3) \le 0$.

1. Problem Description

The problem is to find the solution set of the inequality $(6x + 1)(4x - 3) \le 0$.

2. Solution Steps

First, we find the roots of the equation $(6x + 1)(4x - 3) = 0$. This occurs when $6x + 1 = 0$ or $4x - 3 = 0$.

Solving $6x + 1 = 0$, we have

$6x = -1$

$x = -\frac{1}{6}$

Solving $4x - 3 = 0$, we have

$4x = 3$

$x = \frac{3}{4}$

Now we consider the inequality $(6x + 1)(4x - 3) \le 0$. We have the roots $x = -\frac{1}{6}$ and $x = \frac{3}{4}$. We can test intervals to determine where the inequality holds.

Interval 1: $x < -\frac{1}{6}$. Let's take $x = -1$. Then $(6(-1) + 1)(4(-1) - 3) = (-6 + 1)(-4 - 3) = (-5)(-7) = 35 > 0$.

Interval 2: $-\frac{1}{6} < x < \frac{3}{4}$. Let's take $x = 0$. Then $(6(0) + 1)(4(0) - 3) = (1)(-3) = -3 < 0$.

Interval 3: $x > \frac{3}{4}$. Let's take $x = 1$. Then $(6(1) + 1)(4(1) - 3) = (6 + 1)(4 - 3) = (7)(1) = 7 > 0$.

Since we want $(6x + 1)(4x - 3) \le 0$, we include the roots $x = -\frac{1}{6}$ and $x = \frac{3}{4}$. Therefore, the solution set is $[-\frac{1}{6}, \frac{3}{4}]$.

3. Final Answer

The solution set is $[-\frac{1}{6}, \frac{3}{4}]$. This corresponds to option C.