Solve the following trigonometric equations for their general solutions: a) sin 2x + cos x = 0 b) 6 cos2 x + sin x − 5 = 0

Case 1: $\cos x = 0$
The reference angle for $\cos x = 0$ is $90^\circ$.
The general solution for $\cos x = 0$ is:

$$x = 90^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z}$$

Case 2: $2 \sin x + 1 = 0$

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

The reference angle for $\sin x = \frac{1}{2}$ is $30^\circ$.
Since $\sin x$ is negative, $x$ lies in the 3rd or 4th quadrant.

For the 3rd quadrant:

$$x = 180^\circ + 30^\circ + k \cdot 360^\circ$$ $$x = 210^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

For the 4th quadrant:

$$x = 360^\circ - 30^\circ + k \cdot 360^\circ$$ $$x = 330^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

The general solutions for (a) are:

$$\boxed{x = 90^\circ + k \cdot 180^\circ \quad \text{or} \quad x = 210^\circ + k \cdot 360^\circ \quad \text{or} \quad x = 330^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}}$$

$$6(1 - \sin^2 x) + \sin x - 5 = 0$$ $$6 - 6 \sin^2 x + \sin x - 5 = 0$$ $$-6 \sin^2 x + \sin x + 1 = 0$$

The equation becomes a quadratic equation: $6y^2 - y - 1 = 0$.
Factor the quadratic equation:

$$(3y + 1)(2y - 1) = 0$$

Case 1: $3 \sin x + 1 = 0$

$$3 \sin x = -1$$ $$\sin x = -\frac{1}{3}$$

The reference angle is $\text{ref} = \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ$.
Since $\sin x$ is negative, $x$ lies in the 3rd or 4th quadrant.

For the 3rd quadrant:

$$x = 180^\circ + 19.47^\circ + k \cdot 360^\circ$$ $$x = 199.47^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

For the 4th quadrant:

$$x = 360^\circ - 19.47^\circ + k \cdot 360^\circ$$ $$x = 340.53^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

Case 2: $2 \sin x - 1 = 0$

$$2 \sin x = 1$$ $$\sin x = \frac{1}{2}$$

The reference angle is $\text{ref} = 30^\circ$.
Since $\sin x$ is positive, $x$ lies in the 1st or 2nd quadrant.

For the 1st quadrant:

$$x = 30^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

For the 2nd quadrant:

$$x = 180^\circ - 30^\circ + k \cdot 360^\circ$$ $$x = 150^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

The general solutions for (b) are:

$$\boxed{x = 30^\circ + k \cdot 360^\circ \quad \text{or} \quad x = 150^\circ + k \cdot 360^\circ \quad \text{or} \quad x = 199.47^\circ + k \cdot 360^\circ \quad \text{or} \quad x = 340.53^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}}$$

$$(2 \cos^2 x - 1) - 7 \cos x - 3 = 0$$ $$2 \cos^2 x - 7 \cos x - 4 = 0$$

The equation becomes a quadratic equation: $2y^2 - 7y - 4 = 0$.
Factor the quadratic equation:

$$(2y + 1)(y - 4) = 0$$

Case 1: $2 \cos x + 1 = 0$

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

The reference angle is $\text{ref} = \arccos\left(\frac{1}{2}\right) = 60^\circ$.
Since $\cos x$ is negative, $x$ lies in the 2nd or 3rd quadrant.

For the 2nd quadrant:

$$x = 180^\circ - 60^\circ + k \cdot 360^\circ$$ $$x = 120^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

For the 3rd quadrant:

$$x = 180^\circ + 60^\circ + k \cdot 360^\circ$$ $$x = 240^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

Case 2: $\cos x - 4 = 0$

$$\cos x = 4$$

This case has no solution because the range of $\cos x$ is $[-1, 1]$.

The general solutions for (c) are:

$$\boxed{x = 120^\circ + k \cdot 360^\circ \quad \text{or} \quad x = 240^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}}$$

$$\cos(x - 30^\circ) = \sin(90^\circ - (x - 30^\circ))$$ $$\cos(x - 30^\circ) = \sin(90^\circ - x + 30^\circ)$$ $$\cos(x - 30^\circ) = \sin(120^\circ - x)$$

Apply the general solution for $\sin A = \sin B$:
$A = B + k \cdot 360^\circ$ or $A = 180^\circ - B + k \cdot 360^\circ$.

Case 1: $2x = (120^\circ - x) + k \cdot 360^\circ$

$$2x + x = 120^\circ + k \cdot 360^\circ$$ $$3x = 120^\circ + k \cdot 360^\circ$$ $$x = \frac{120^\circ}{3} + \frac{k \cdot 360^\circ}{3}$$ $$x = 40^\circ + k \cdot 120^\circ, \quad k \in \mathbb{Z}$$

Case 2: $2x = 180^\circ - (120^\circ - x) + k \cdot 360^\circ$

$$2x = 180^\circ - 120^\circ + x + k \cdot 360^\circ$$ $$2x = 60^\circ + x + k \cdot 360^\circ$$ $$2x - x = 60^\circ + k \cdot 360^\circ$$ $$x = 60^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$

The general solutions for (d) are:

$$\boxed{x = 40^\circ + k \cdot 120^\circ \quad \text{or} \quad x = 60^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}}$$

$$\sqrt{3} \cos x = 4 \sin x - \sin x$$ $$\sqrt{3} \cos x = 3 \sin x$$

$$\frac{\sqrt{3}}{3} = \frac{\sin x}{\cos x}$$ $$\tan x = \frac{\sqrt{3}}{3}$$

The reference angle for $\tan x = \frac{\sqrt{3}}{3}$ is $30^\circ$.
The general solution for $\tan x = \tan \alpha$ is $x = \alpha + k \cdot 180^\circ$.

$$x = 30^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z}$$

The general solution for (e) is:

$$\boxed{x = 30^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z}}$$