Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$ 
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$.

$4\cos x - 3 \sin x = k\sin(x - \alpha)$
=> $k(\sin x \cos \alpha - \cos x \sin \alpha)$
=> $k\cos \alpha \sin x - \sin \alpha \cos x$
Equating coefficients:
$k\cos \alpha = - 3$
$k\sin \alpha = 4$
$k = \sqrt{(-3)^2 + 4^2} = 5$
$\alpha$ is in the fourth quarter because $ \cos \alpha$ is positive and $\sin \alpha$ is negative.
$\alpha = \arctan\frac{4}{-3} = -53.1$
$\alpha = 360 - 53.1 = 306.9$
$4\cos x - 3 \sin x = 5\sin(x - 306.9)$
But the answer is $4\cos x - 3 \sin x = 5\sin(x - 233.1)$

$$4\cos x - 3 \sin x = k\sin(x - \alpha)$$
$$= k(\sin x \cos \alpha - \cos x \sin \alpha)$$
$$=k\cos \alpha \sin x - k\sin \alpha \cos x$$
Hence:
$$k\cos \alpha \sin x = - 3 \sin x$$
$$- k\sin \alpha \cos x = 4\cos x$$
so:
$$k\cos \alpha = -3$$
$$\color{red}{k\sin \alpha = - 4}$$
