Section A: Pure Mathematics I Prove the trigonometric identity: sin(3θ) ------- sin θ -- cos(3θ) ------- cos θ = 2

We know that $\cos(2\theta) = 1 - 2\sin^2\theta$. Substitute this into the given equation.

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

$$1 - 2\sin^2\theta + 3\sin\theta - 2 = 0$$ $$-2\sin^2\theta + 3\sin\theta - 1 = 0$$

Multiply by $-1$ to make the leading coefficient positive:

$$2\sin^2\theta - 3\sin\theta + 1 = 0$$

Let $x = \sin\theta$. The equation becomes $2x^2 - 3x + 1 = 0$.
Factor the quadratic equation:

$$(2x - 1)(x - 1) = 0$$

This gives two possible values for $x$:

$$2x - 1 = 0 \implies x = \frac{1}{2}$$ $$x - 1 = 0 \implies x = 1$$

So, we have $\sin\theta = \frac{1}{2}$ or $\sin\theta = 1$.

The only angle in this interval for which $\sin\theta = 1$ is:

$$\theta = 90^\circ$$

The principal value (in the first quadrant) for which $\sin\theta = \frac{1}{2}$ is:

$$\theta_1 = 30^\circ$$

Since $\sin\theta$ is also positive in the second quadrant, the other solution in the interval is:

$$\theta_2 = 180^\circ - 30^\circ = 150^\circ$$

The solutions for $\theta$ in the interval $0^\circ \le \theta \le 360^\circ$ are $30^\circ$, $90^\circ$, and $150^\circ$.

The solutions are $30^\circ, 90^\circ, 150^\circ$.