Without using a calculator, solve the inequality $\dfrac{6x-4}{x-3} \le 1-2x$, $x \ne 3$.

1. Solve the rational inequality from the previous part to obtain the required solution intervals.

$\frac{6x-4}{x-3} \le 1-2x$\n$\frac{6x-4 - (1-2x)(x-3)}{x-3} \le 0$\n$\frac{6x-4 - (-2x^2+7x-3)}{x-3} \le 0$\n$\frac{2x^2-x-1}{x-3} \le 0$\n$\frac{(2x+1)(x-1)}{x-3} \le 0$\nFrom a sign test on the critical values $-\frac{1}{2}$, $1$, and $3$, the valid intervals are $x \le -\frac{1}{2}$ or $1 \le x < 3$.

2. Manipulate the given trigonometric inequality to match the form of the solved rational inequality.

$\frac{6-4\csc\theta}{1-3\csc\theta} \le 1-2\sin\theta$\nSubstitute $\csc\theta = \frac{1}{\sin\theta}$:\n$\frac{6 - \frac{4}{\sin\theta}}{1 - \frac{3}{\sin\theta}} \le 1-2\sin\theta$\nMultiply the numerator and denominator of the fraction by $\sin\theta$:\n$\frac{6\sin\theta-4}{\sin\theta-3} \le 1-2\sin\theta$\nThis perfectly matches the inequality in the first part with $x = \sin\theta$.

3. Substitute $x = \sin\theta$ into the solution intervals and deduce the exact values for $\theta$.

Substituting $x = \sin\theta$ into the solution intervals gives:\n$\sin\theta \le -\frac{1}{2} \quad \text{or} \quad 1 \le \sin\theta < 3$\nFor $1 \le \sin\theta < 3$, since the maximum possible value of $\sin\theta$ is $1$, we must have exactly $\sin\theta = 1$.\nFor the domain $0 < \theta < 2\pi$, this yields $\theta = \frac{\pi}{2}$.\n\nFor $\sin\theta \le -\frac{1}{2}$, the basic (reference) angle is $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$.\nSince $\sin\theta$ is negative, $\theta$ must lie in the 3rd or 4th quadrant:\n$\pi + \frac{\pi}{6} \le \theta \le 2\pi - \frac{\pi}{6}$\n$\frac{7\pi}{6} \le \theta \le \frac{11\pi}{6}$