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!

This is just my way of saying what everyone else has already said.
\begin{align}
   R &=\sqrt{1^2+5^2}=\sqrt{26} \\
   \cos \alpha &= \dfrac{5}{\sqrt{26}} \\
   \sin \alpha &= \dfrac{1}{\sqrt{26}} \\
   \alpha &\approx 11.31^\circ \\
\hline
   5\cos x - 1 \sin x &= 4 \\
   \cos x \cos \alpha - \sin x \sin \alpha &=  \dfrac{4}{\sqrt{26}} \\
   \cos(x + \alpha) &= \dfrac{4}{\sqrt{26}} \\
   x + 11.31^\circ &\approx \pm 38.33^\circ + n 360^\circ \\
\hline
   x &\approx 38.33^\circ - 11.31^\circ = 27.02^\circ \\
   x &\approx -38.33^\circ - 11.31^\circ + 360^\circ = 310.36^\circ
\end{align}
