The problem asks to solve the trigonometric inequality $2\sin^2\theta \ge 3\cos\theta$.

1. Problem Description

The problem asks to solve the trigonometric inequality $2\sin^2\theta \ge 3\cos\theta$.

2. Solution Steps

First, we rewrite the inequality using the identity $\sin^2\theta + \cos^2\theta = 1$, so $\sin^2\theta = 1 - \cos^2\theta$.

Substituting this into the inequality, we get:

$2(1 - \cos^2\theta) \ge 3\cos\theta$

$2 - 2\cos^2\theta \ge 3\cos\theta$

$0 \ge 2\cos^2\theta + 3\cos\theta - 2$

$2\cos^2\theta + 3\cos\theta - 2 \le 0$

We can factor the quadratic expression in terms of $\cos\theta$ as follows:

$(2\cos\theta - 1)(\cos\theta + 2) \le 0$

Since $-1 \le \cos\theta \le 1$, $\cos\theta + 2$ is always positive. Thus, we only need to consider the factor $(2\cos\theta - 1)$.

For the inequality to hold, we must have $2\cos\theta - 1 \le 0$, which means

$2\cos\theta \le 1$

$\cos\theta \le \frac{1}{2}$

We need to find the values of $\theta$ for which $\cos\theta \le \frac{1}{2}$.

The reference angle for $\cos\theta = \frac{1}{2}$ is $\theta = \frac{\pi}{3}$.

Since $\cos\theta$ is positive in the first and fourth quadrants and negative in the second and third quadrants, $\cos\theta \le \frac{1}{2}$ when $\theta$ is in the second and third quadrants. The range of angles satisfying $\cos\theta \le \frac{1}{2}$ is $\frac{\pi}{3} \le \theta \le 2\pi - \frac{\pi}{3}$, which simplifies to $\frac{\pi}{3} \le \theta \le \frac{5\pi}{3}$.

$\cos(\frac{\pi}{3}) = \frac{1}{2}$. Therefore, $\theta = \frac{\pi}{3}$ is the boundary.

$\cos(\frac{5\pi}{3}) = \cos(2\pi - \frac{\pi}{3}) = \cos(-\frac{\pi}{3}) = \frac{1}{2}$. Therefore, $\theta = \frac{5\pi}{3}$ is the boundary.

3. Final Answer

$\frac{\pi}{3} \le \theta \le \frac{5\pi}{3}$