Problem source

Show that if $\, \cos(x - \alpha) = \cos \beta \,$  
then either $\, \tan x = \tan ( \alpha + \beta)\,$ or
$\; \tan x = \tan ( \alpha - \beta)\,$. 
By choosing suitable values of $x$, $\alpha$ and $\beta\,$,
give an example to show that if 
$\,\tan x = \tan ( \alpha + \beta)\,$, 
then $\,\cos(x - \alpha) \, $ need not equal $ \cos \beta \,$.
Let $\omega$ be the acute angle such that $\tan \omega = \frac 43\,$.
\begin{questionparts}
\item For $0 \le x \le 2 \pi$, solve the equation
\[
\cos x -7 \sin x = 5
\]
giving both solutions in terms of $\omega\,$.
\item  For $0 \le x \le 2 \pi$, solve the equation
\[
2\cos x + 11 \sin x = 10
\]
showing that one solution is twice the other and giving both in terms of $\omega\,$.
\end{questionparts}