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.

The general method of solving $A\cos x+B\sin x=C$ when $A^2+B^2\ne 0$ is to take any $y$ such that $$(1)\quad A/\sqrt {A^2+B^2}=\cos y\;\land \;B/\sqrt {A^2+B^2}=\sin y.$$  $$\text {Then } C/\sqrt {A^2+B^2}=\cos y \cos x+\sin y \sin x =\cos (x-y).$$ So let  $x=y+z$ for any $z$ such that $\cos z=C/\sqrt {A^2+B^2}.$
If we are strictly in the real numbers we require $|C/\sqrt {A^2+B^2}|\leq 1......$ Let $y_0$ satisfy (1) with $y_0\in [0,\pi].$  Let $z_0=\cos^{-1}(C/\sqrt {A^2+B^2})\in [0,\pi].$ Then $x=2\pi n \pm (y_0+z_0)$ for some (any) integer $n.$
