<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] \begin{eqnarray}
5\cos x - \sin x &=& 4\\
\frac{5}{\sqrt{26}}\cos x-\frac{1}{\sqrt{26}}\sin x&=&\frac{4}{\sqrt{26}}\\
\cos\alpha\cos x-\sin\alpha\sin x&=&\frac{4}{\sqrt{26}}\\
\end{eqnarray}
Giving
$$\cos(x+\alpha)=\frac{4}{\sqrt{26}}$$
So either
$$ x+\alpha=\arccos\left(\frac{4}{\sqrt{26}}\right)=38.33^\circ $$
or
$$ x+\alpha=360^\circ-\arccos\left(\frac{4}{\sqrt{26}}\right)=321.67^\circ $$
with
$$\alpha=\arccos\left(\frac{5}{\sqrt{26}}\right)=11.31$$
So you get that either $x=27.02^\circ$ or $x=310.36^\circ$.
</s>