The problem asks us to solve several trigonometric equations in $\mathbb{R}$ and represent the solutions on the unit circle. We will solve the first two equations: a) $3\cos x - \sqrt{3}\sin x + \sqrt{6} = 0$ b) $\cos 2x - \sin 2x = -1$

1. Problem Description

The problem asks us to solve several trigonometric equations in $\mathbb{R}$ and represent the solutions on the unit circle. We will solve the first two equations:

a) $3\cos x - \sqrt{3}\sin x + \sqrt{6} = 0$

b) $\cos 2x - \sin 2x = -1$

2. Solution Steps

a) $3\cos x - \sqrt{3}\sin x + \sqrt{6} = 0$

We can rewrite the equation as:

$\sqrt{3}\sin x - 3\cos x = \sqrt{6}$

Divide by $\sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3+9} = \sqrt{12} = 2\sqrt{3}$:

$\frac{\sqrt{3}}{2\sqrt{3}}\sin x - \frac{3}{2\sqrt{3}}\cos x = \frac{\sqrt{6}}{2\sqrt{3}}$

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

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

$\sin(x - \frac{\pi}{3}) = \frac{\sqrt{2}}{2}$

So $x - \frac{\pi}{3} = \frac{\pi}{4} + 2k\pi$ or $x - \frac{\pi}{3} = \frac{3\pi}{4} + 2k\pi$ for $k \in \mathbb{Z}$.

$x = \frac{\pi}{4} + \frac{\pi}{3} + 2k\pi = \frac{3\pi + 4\pi}{12} + 2k\pi = \frac{7\pi}{12} + 2k\pi$

$x = \frac{3\pi}{4} + \frac{\pi}{3} + 2k\pi = \frac{9\pi + 4\pi}{12} + 2k\pi = \frac{13\pi}{12} + 2k\pi$

b) $\cos 2x - \sin 2x = -1$

We can rewrite the equation as:

$\cos 2x - \sin 2x + 1 = 0$

We can divide by $\sqrt{1^2 + (-1)^2} = \sqrt{2}$:

$\frac{1}{\sqrt{2}}\cos 2x - \frac{1}{\sqrt{2}}\sin 2x = - \frac{1}{\sqrt{2}}$

$\cos(\frac{\pi}{4})\cos 2x - \sin(\frac{\pi}{4})\sin 2x = -\frac{\sqrt{2}}{2}$

$\cos(2x + \frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$

So $2x + \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi$ or $2x + \frac{\pi}{4} = \frac{5\pi}{4} + 2k\pi$ for $k \in \mathbb{Z}$.

$2x = \frac{3\pi}{4} - \frac{\pi}{4} + 2k\pi = \frac{2\pi}{4} + 2k\pi = \frac{\pi}{2} + 2k\pi$

$x = \frac{\pi}{4} + k\pi$

$2x = \frac{5\pi}{4} - \frac{\pi}{4} + 2k\pi = \frac{4\pi}{4} + 2k\pi = \pi + 2k\pi$

$x = \frac{\pi}{2} + k\pi$

Therefore, $x = \frac{\pi}{4} + k\pi$ or $x = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$

3. Final Answer

a) $x = \frac{7\pi}{12} + 2k\pi$ or $x = \frac{13\pi}{12} + 2k\pi$, $k \in \mathbb{Z}$

b) $x = \frac{\pi}{4} + k\pi$ or $x = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$