Solve the equation $3 \cos 2x = \sin x + 2$, for $0^\circ \le x \le 360^\circ$. Find all the angles between 0 and 5 which satisfy the equation $\sin 2x + 3 \cos^2 x = 0$.

1. Express the equation in terms of $\sin x$ using the double angle formula

$3(1 - 2\sin^2 x) = \sin x + 2$

2. Rearrange into a quadratic equation in $\sin x$ and solve

$3 - 6\sin^2 x = \sin x + 2$
$6\sin^2 x + \sin x - 1 = 0$
$(3\sin x - 1)(2\sin x + 1) = 0$
$\sin x = \frac{1}{3} \quad \text{or} \quad \sin x = -\frac{1}{2}$

3. Find the solutions for $\sin x = \frac{1}{3}$

Basic angle $\alpha = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.471^\circ$
Since $\sin x > 0$, $x$ is in the 1st and 2nd quadrants:
$x = 19.471^\circ \quad \text{or} \quad 180^\circ - 19.471^\circ = 160.529^\circ$
$x \approx 19.5^\circ, 160.5^\circ \text{ (to 1 d.p.)}$

4. Find the solutions for $\sin x = -\frac{1}{2}$

Basic angle $\alpha = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$
Since $\sin x < 0$, $x$ is in the 3rd and 4th quadrants:
$x = 180^\circ + 30^\circ = 210^\circ \quad \text{or} \quad 360^\circ - 30^\circ = 330^\circ$

1. Apply the double angle identity for sine

$\sin 2x + 3 \cos^2 x = 0$
$2 \sin x \cos x + 3 \cos^2 x = 0$

2. Factorise the equation

$\cos x (2 \sin x + 3 \cos x) = 0$

3. Solve for the first factor within the given range

$\cos x = 0$
Since $0 \le x \le 5$ (in radians):
$x = \frac{\pi}{2}, \frac{3\pi}{2}$
$x \approx 1.5708, 4.7124$

4. Solve for the second factor within the given range

$2 \sin x + 3 \cos x = 0$
$2 \sin x = -3 \cos x$
$\tan x = -1.5$
Basic angle $\alpha = \tan^{-1}(1.5) \approx 0.98279$
Since $\tan x$ is negative, $x$ is in the 2nd and 4th quadrants.
$x = \pi - 0.98279 \approx 2.1588$
$x = 2\pi - 0.98279 \approx 5.3004$ (rejected as $5.3004 > 5$)