12

1. Apply the double angle identity for sine

$0.5(2\sin x \cos x) = 3\sin x$
$\sin x \cos x = 3\sin x$

2. Rearrange and factorise the equation

$\sin x \cos x - 3\sin x = 0$
$\sin x (\cos x - 3) = 0$

3. Solve for $x$ in the given interval

Either $\sin x = 0$ or $\cos x - 3 = 0$

For $\cos x = 3$, there are no real solutions since $-1 \leqslant \cos x \leqslant 1$.

For $\sin x = 0$, the solutions in the interval $0 < x < 2\pi$ are:
$x = \pi$

1. Express the equation in terms of sine and find the basic angle.

$\csc\left(\frac{\theta}{2}-40^\circ\right) = -5$
$\sin\left(\frac{\theta}{2}-40^\circ\right) = -\frac{1}{5} = -0.2$
$\text{Basic angle } \alpha = \sin^{-1}(0.2) \approx 11.537^\circ$

2. Determine the valid range for the angle argument.

Given $0^\circ < \theta < 180^\circ$:
$0^\circ < \frac{\theta}{2} < 90^\circ$
$-40^\circ < \frac{\theta}{2}-40^\circ < 50^\circ$

3. Solve for the angle in the correct quadrant and find $\theta$.

Since sine is negative and the angle is between $-40^\circ$ and $50^\circ$, the angle must be in the 4th quadrant (negative angle):
$\frac{\theta}{2}-40^\circ = -11.537^\circ$
$\frac{\theta}{2} = 28.463^\circ$
$\theta = 56.926^\circ \approx 56.9^\circ \text{ (to 1 d.p.)}$