Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin x$$
equating coefficients:
$k\cos\alpha = 8$
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$? I find this really confusing
$k = \sqrt{8^2 + 6^2} = 10$
$\alpha$ is in the 1st quadrant where both sin and cos are positive.
$\alpha = \arctan\frac{6}8 = 36.8^{\circ}$
$\therefore10\cos(x - 36.8) = 5$
I have a maximum and minimum value of 10
From here I do not know how to finish solving for x.


$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$? 

It is the latter.
So, having $\tan\alpha=-6/8=-3/4$ gives $\alpha=\arctan(-3/4)\approx -36.9^\circ$.
Hence, for $n\in\mathbb Z$,
$$\begin{align}10\cos(x-(-36.9^\circ))=5&\Rightarrow \cos(x+36.9^\circ)=1/2\\&\Rightarrow x+36.9^\circ=\pm 60^\circ+360^\circ n\\&\Rightarrow x=-36.9^\circ\pm 60^\circ+360^\circ n\\&\Rightarrow x=23.1^\circ,\quad 263.1^\circ\end{align}$$
since $0^\circ\le x\le 360^\circ$.
