Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And then assume:
$$
  \frac{6}{\sqrt{51}}= \cos \psi ; \frac{5}{\sqrt{51}}= \sin\psi ; 
$$
$$
  \cos \psi \cos x - \sin \psi \sin x = \cos (x+ \psi) = \cos(x + \arccos ( \frac{6}{\sqrt{51}}))
$$
$$
x + \arccos\left(\frac{6}{\sqrt{51}}\right) = \arcsin\left( \frac{8}{\sqrt{51}}\right)
$$
$$
x \approx 12^\circ
$$
But answer is:
$$
 -\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z
$$

We have $$6 \cos x - 5\cos x = \cos x$$
But, a quick look at the graph of the cosine function shows us that it is bounded between $-1$ and $1$, so $\cos x = 8$ has no solutions. $\square$

Assuming your original equation was $6 \cos x - 5\sin x = 8$ as your body would suggest, we can represent this in the form $$\sqrt{61} \sin \left(x - \arctan\left(\frac{5}{6}\right)\right) = 8$$
using the same method you did in your question, can you take it from there?
Edit: This still doesn't make sense, we have $$\max(6 \cos x - 5\sin x) = \sqrt{61} < 8$$ So there are still no solutions. 
