Addma compound and multiple angles solve simply

$$\cos(\theta + 30^\circ) = \cos\theta \cos 30^\circ - \sin\theta \sin 30^\circ$$

Substitute the known values $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$:

$$\cos(\theta + 30^\circ) = \frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta$$

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

Divide both sides by $\cos\theta$ and $\frac{3}{2}$:

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

The reference angle for $\tan\theta = \frac{\sqrt{3}}{3}$ is $30^\circ$.
Since $\tan\theta$ is positive, $\theta$ lies in Quadrant I and Quadrant III.
In Quadrant I: $\theta = 30^\circ$.
In Quadrant III: $\theta = 180^\circ + 30^\circ = 210^\circ$.
The solutions are $\boxed{\theta = 30^\circ, 210^\circ}$.

5b) Solve $\sqrt{2}\cos(\theta - 45^\circ) = 4\sin\theta$ for $0^\circ \le \theta < 360^\circ$.

$$\cos(\theta - 45^\circ) = \cos\theta \cos 45^\circ + \sin\theta \sin 45^\circ$$

Substitute the known values $\cos 45^\circ = \frac{\sqrt{2}}{2}$ and $\sin 45^\circ = \frac{\sqrt{2}}{2}$:

$$\cos(\theta - 45^\circ) = \frac{\sqrt{2}}{2}\cos\theta + \frac{\sqrt{2}}{2}\sin\theta$$

$$\sqrt{2}\left(\frac{\sqrt{2}}{2}\cos\theta + \frac{\sqrt{2}}{2}\sin\theta\right) = 4\sin\theta$$ $$\left(\frac{2}{2}\cos\theta + \frac{2}{2}\sin\theta\right) = 4\sin\theta$$ $$\cos\theta + \sin\theta = 4\sin\theta$$

$$\cos\theta = 4\sin\theta - \sin\theta$$ $$\cos\theta = 3\sin\theta$$

Divide both sides by $\cos\theta$:

$$1 = 3\frac{\sin\theta}{\cos\theta}$$ $$\tan\theta = \frac{1}{3}$$

The reference angle for $\tan\theta = \frac{1}{3}$ is $\arctan\left(\frac{1}{3}\right) \approx 18.43^\circ$.
Since $\tan\theta$ is positive, $\theta$ lies in Quadrant I and Quadrant III.
In Quadrant I: $\theta \approx 18.43^\circ$.
In Quadrant III: $\theta = 180^\circ + 18.43^\circ = 198.43^\circ$.
The solutions are $\boxed{\theta \approx 18.43^\circ, 198.43^\circ}$.

6a) Solve $\sin 50^\circ \cos x + \cos 50^\circ \sin x = 1$ for $0^\circ \le x < 360^\circ$.

The left side of the equation matches this form with $A = 50^\circ$ and $B = x$.
So, the equation can be rewritten as:

$$\sin(50^\circ + x) = 1$$

The sine function is equal to 1 when the angle is $90^\circ$ plus any multiple of $360^\circ$.

$$50^\circ + x = 90^\circ + n \cdot 360^\circ$$

$$x = 90^\circ - 50^\circ + n \cdot 360^\circ$$ $$x = 40^\circ + n \cdot 360^\circ$$

For $n=0$: $x = 40^\circ$.
For other integer values of $n$, $x$ will be outside the given domain.
The solution is $\boxed{x = 40^\circ}$.

6b) Solve $\tan x - \tan 35^\circ = 1 + \tan x \tan 35^\circ$ for $0^\circ \le x < 360^\circ$.

Divide both sides by $(1 + \tan x \tan 35^\circ)$:

$$\frac{\tan x - \tan 35^\circ}{1 + \tan x \tan 35^\circ} = 1$$

So, the equation becomes:

$$\tan(x - 35^\circ) = 1$$

The tangent function is equal to 1 when the angle is $45^\circ$ plus any multiple of $180^\circ$.

$$x - 35^\circ = 45^\circ + n \cdot 180^\circ$$

$$x = 45^\circ + 35^\circ + n \cdot 180^\circ$$ $$x = 80^\circ + n \cdot 180^\circ$$

For $n=0$: $x = 80^\circ$.
For $n=1$: $x = 80^\circ + 180^\circ = 260^\circ$.
For other integer values of $n$, $x$ will be outside the given domain.
The solutions are $\boxed{x = 80^\circ, 260^\circ}$.