<s>[INST] All the roots of $5\cos x - \sin x = 4$ in the interval $0^{\circ} \leq x \leq 360^{\circ}$? This is a problem that I stumbled upon in one of my books.
Representing $5\cos x - \sin x$ in the form $R\cos(x + \alpha)$ (as demanded by the question):
$
\rightarrow R = \sqrt{5^2 + ({-}1)^2} = \sqrt{26}\\
\rightarrow R\cos x \cos \alpha - R\sin x \sin \alpha = 5\cos x - \sin x \\
\rightarrow ➊\hspace{0.25cm}5 = \sqrt{26}\cos \alpha \\
\rightarrow ➋\hspace{0.25cm}{-}1 = \sqrt{26}\sin \alpha \\
$ 
So here are my question(s): $\\$ 
• Why is only ➊ working out?
• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?
• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $\cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!
 [/INST] It seems that you know why $R=\sqrt{26}$,
Hence $5\cos x - \sin x$ can be expressed as $\sqrt{26}(\frac{5}{\sqrt{26}}\cos x - \frac{1}{\sqrt{26}}\sin x)$
$$\text{Let, } \cos\alpha = \frac{5}{\sqrt{26}} \text{ and } \sin\alpha = \frac{1}{\sqrt{26}}$$
Hence, $5\cos x - \sin x$ can now be expressed as $\sqrt{26}\cos (x + \alpha)$.
To solve your initial equation, we can now write it as,
$$\sqrt{26}\cos (x + \alpha) =4$$
$$\cos (x + \alpha) = \frac{4}{\sqrt{26}}$$
$$x +\alpha = \cos^{-1} \frac{4}{\sqrt{26}}$$
If $\cos\alpha = \frac{5}{\sqrt{26}}$, then $\alpha = \cos^{-1} \frac{5}{\sqrt{26}}$,
$$x = \cos^{-1} \frac{4}{\sqrt{26}} - \alpha \rightarrow x = \cos^{-1} \frac{4}{\sqrt{26}} - \cos^{-1} \frac{5}{\sqrt{26}}$$
$$\therefore x = 27^o \text{ or } 321.40^o$$
</s>