We are asked to solve the following trigonometric equations in $R$ and represent the solutions on the unit circle: a) $3\cos x - \sqrt{3}\sin x + \sqrt{6} = 0$ b) $\cos 2x - \sin 2x = -1$ c) $\cos^2(2x-\frac{\pi}{3}) - \cos^2(x+\frac{\pi}{4}) = 0$ d) $\sin 2x - 2\sin x \cos(x+\frac{\pi}{3}) = 0$

1. Problem Description

We are asked to solve the following trigonometric equations in $R$ and represent the solutions on the unit circle:

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

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

c) $\cos^2(2x-\frac{\pi}{3}) - \cos^2(x+\frac{\pi}{4}) = 0$

d) $\sin 2x - 2\sin x \cos(x+\frac{\pi}{3}) = 0$

2. Solution Steps

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

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

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

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

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

$\cos(\frac{\pi}{6})\cos x - \sin(\frac{\pi}{6})\sin x = -\frac{\sqrt{2}}{2}$

$\cos(x + \frac{\pi}{6}) = -\frac{\sqrt{2}}{2}$

$x + \frac{\pi}{6} = \frac{3\pi}{4} + 2k\pi$ or $x + \frac{\pi}{6} = -\frac{3\pi}{4} + 2\pi + 2k\pi = \frac{5\pi}{4} + 2k\pi$, $k \in Z$

$x = \frac{3\pi}{4} - \frac{\pi}{6} + 2k\pi = \frac{9\pi - 2\pi}{12} + 2k\pi = \frac{7\pi}{12} + 2k\pi$

or

$x = \frac{5\pi}{4} - \frac{\pi}{6} + 2k\pi = \frac{15\pi - 2\pi}{12} + 2k\pi = \frac{13\pi}{12} + 2k\pi$

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

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

Divide by $\sqrt{1^2 + (-1)^2} = \sqrt{2}$

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

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

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

$2x + \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi$ or $2x + \frac{\pi}{4} = \frac{5\pi}{4} + 2k\pi$, $k \in Z$

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

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

or

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

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

c) $\cos^2(2x-\frac{\pi}{3}) - \cos^2(x+\frac{\pi}{4}) = 0$

Using $a^2-b^2=(a-b)(a+b)$:

$(\cos(2x-\frac{\pi}{3}) - \cos(x+\frac{\pi}{4}))(\cos(2x-\frac{\pi}{3}) + \cos(x+\frac{\pi}{4})) = 0$

Using the cosine sum-to-product formulas:

$\cos A - \cos B = -2 \sin(\frac{A+B}{2}) \sin(\frac{A-B}{2})$

$\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$

$-4\sin(\frac{2x-\frac{\pi}{3}+x+\frac{\pi}{4}}{2}) \sin(\frac{2x-\frac{\pi}{3}-x-\frac{\pi}{4}}{2}) \cos(\frac{2x-\frac{\pi}{3}+x+\frac{\pi}{4}}{2}) \cos(\frac{2x-\frac{\pi}{3}-x-\frac{\pi}{4}}{2}) = 0$

$-4\sin(\frac{3x-\frac{\pi}{12}}{2}) \sin(\frac{x-\frac{7\pi}{12}}{2}) \cos(\frac{3x-\frac{\pi}{12}}{2}) \cos(\frac{x-\frac{7\pi}{12}}{2}) = 0$

$- \sin(3x-\frac{\pi}{12}) \sin(x-\frac{7\pi}{12}) = 0$

$\sin(3x-\frac{\pi}{12}) = 0$ or $\sin(x-\frac{7\pi}{12}) = 0$

$3x-\frac{\pi}{12} = k\pi$ or $x-\frac{7\pi}{12} = k\pi$, $k \in Z$

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

$x = \frac{k\pi}{3} + \frac{\pi}{36}$ or $x = k\pi + \frac{7\pi}{12}$

d) $\sin 2x - 2\sin x \cos(x+\frac{\pi}{3}) = 0$

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

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

$\sin x = 0$ or $\cos x = \cos(x+\frac{\pi}{3})$

$x = k\pi$ or $x = x + \frac{\pi}{3} + 2k\pi$ or $x = -(x+\frac{\pi}{3}) + 2k\pi$

$\sin x = 0$

$x = k\pi$

$\cos x = \cos(x+\frac{\pi}{3})$ has no solutions.

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

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

$x = -\frac{\pi}{6} + k\pi$

So, $x = k\pi$ or $x = -\frac{\pi}{6} + k\pi$

3. Final Answer

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

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

c) $x = \frac{k\pi}{3} + \frac{\pi}{36}$ or $x = k\pi + \frac{7\pi}{12}$, $k \in Z$

d) $x = k\pi$ or $x = -\frac{\pi}{6} + k\pi$, $k \in Z$