We are asked to solve three equations and one inequality for $x$ in the interval $0 \le x < 2\pi$. The problems are: (3) $\sin 2x = 2 \cos x$ (4) $\cos 2x + \cos x \ge 0$ (5) $\cos x + \cos 3x = 0$ (6) $\cos x + \sqrt{3}(\sin x + 1) \ge 0$

1. Problem Description

We are asked to solve three equations and one inequality for $x$ in the interval $0 \le x < 2\pi$. The problems are:

(3) $\sin 2x = 2 \cos x$

(4) $\cos 2x + \cos x \ge 0$

(5) $\cos x + \cos 3x = 0$

(6) $\cos x + \sqrt{3}(\sin x + 1) \ge 0$

2. Solution Steps

(3) $\sin 2x = 2 \cos x$

Using the double angle formula, $\sin 2x = 2 \sin x \cos x$.

$2 \sin x \cos x = 2 \cos x$

$2 \sin x \cos x - 2 \cos x = 0$

$2 \cos x (\sin x - 1) = 0$

Thus, $\cos x = 0$ or $\sin x = 1$.

If $\cos x = 0$, then $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

If $\sin x = 1$, then $x = \frac{\pi}{2}$.

Therefore, the solutions are $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

(4) $\cos 2x + \cos x \ge 0$

Using the double angle formula, $\cos 2x = 2 \cos^2 x - 1$.

$2 \cos^2 x - 1 + \cos x \ge 0$

$2 \cos^2 x + \cos x - 1 \ge 0$

$(2 \cos x - 1)(\cos x + 1) \ge 0$

We have two cases:

Case 1: $2 \cos x - 1 \ge 0$ and $\cos x + 1 \ge 0$

$\cos x \ge \frac{1}{2}$ and $\cos x \ge -1$. Since $\cos x$ is always greater or equal to -1, we only need to consider $\cos x \ge \frac{1}{2}$. The solutions for this are $0 \le x \le \frac{\pi}{3}$ and $\frac{5\pi}{3} \le x < 2\pi$.

Case 2: $2 \cos x - 1 \le 0$ and $\cos x + 1 \le 0$

$\cos x \le \frac{1}{2}$ and $\cos x \le -1$. Then $\cos x = -1$, so $x = \pi$.

Combining the cases, we have $0 \le x \le \frac{\pi}{3}$, $x = \pi$, and $\frac{5\pi}{3} \le x < 2\pi$.

(5) $\cos x + \cos 3x = 0$

Using the sum-to-product formula, $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$.

$\cos x + \cos 3x = 2 \cos \frac{x+3x}{2} \cos \frac{x-3x}{2} = 2 \cos 2x \cos (-x) = 2 \cos 2x \cos x = 0$.

So $\cos 2x = 0$ or $\cos x = 0$.

If $\cos 2x = 0$, then $2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$. Thus, $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.

If $\cos x = 0$, then $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

Therefore, the solutions are $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{2}$.

(6) $\cos x + \sqrt{3}(\sin x + 1) \ge 0$

$\cos x + \sqrt{3} \sin x + \sqrt{3} \ge 0$

$\cos x + \sqrt{3} \sin x \ge - \sqrt{3}$

Divide by 2:

$\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \ge - \frac{\sqrt{3}}{2}$

$\cos \frac{\pi}{3} \cos x + \sin \frac{\pi}{3} \sin x \ge - \frac{\sqrt{3}}{2}$

$\cos(x - \frac{\pi}{3}) \ge - \frac{\sqrt{3}}{2}$

The interval where $\cos \theta \ge -\frac{\sqrt{3}}{2}$ is $-\frac{5\pi}{6} \le \theta \le \frac{5\pi}{6}$.

Thus, $-\frac{5\pi}{6} \le x - \frac{\pi}{3} \le \frac{5\pi}{6}$.

Adding $\frac{\pi}{3}$ to all sides:

$-\frac{5\pi}{6} + \frac{\pi}{3} \le x \le \frac{5\pi}{6} + \frac{\pi}{3}$

$-\frac{5\pi}{6} + \frac{2\pi}{6} \le x \le \frac{5\pi}{6} + \frac{2\pi}{6}$

$-\frac{3\pi}{6} \le x \le \frac{7\pi}{6}$

$-\frac{\pi}{2} \le x \le \frac{7\pi}{6}$

Since $0 \le x < 2\pi$, we have $0 \le x \le \frac{7\pi}{6}$.

3. Final Answer

(3) $x = \frac{\pi}{2}, \frac{3\pi}{2}$

(4) $0 \le x \le \frac{\pi}{3}, x = \pi, \frac{5\pi}{3} \le x < 2\pi$

(5) $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{2}$

(6) $0 \le x \le \frac{7\pi}{6}$