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:
$$
 -\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z
$$
 A: 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. 
A: $$6\cos { x } -5\sin { x } =8\\ 6\cos ^{ 2 }{ \frac { x }{ 2 } -6\sin ^{ 2 }{ \frac { x }{ 2 } -10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } =8 } \cos ^{ 2 }{ \frac { x }{ 2 }  } +8\sin ^{ 2 }{ \frac { x }{ 2 }  }  }  }  } \\ 14\sin ^{ 2 }{ \frac { x }{ 2 }  } +10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 }  }  } +2\cos ^{ 2 }{ \frac { x }{ 2 }  } =0\\ \cos ^{ 2 }{ \frac { x }{ 2 }  } \neq 0$$
$$14\tan ^{ 2 }{ \frac { x }{ 2 } +10\tan { \frac { x }{ 2 } +2=0 }  } \\ \tan { \frac { x }{ 2 } =a } \\ 7{ a }^{ 2 }+5a+1=0\\ $$
can you take from here?
A: By the Cauchy-Schwarz inequality:
$$ \left( 6\cos x-5\sin x\right)^2 \leq 36+25 = 61<64 $$
so it is not possible that $6\cos x-5\sin x$ equals $8$ for some real $x$.
A: Edit: In general, the maximum value of $a\cos A+b\sin A$ is $\sqrt{a^2+b^2}$ 
Hence the maximum value of $6\cos x-5\sin x$ is $\sqrt{6^2+(-5)^2}=\sqrt{61}$ 
But $RHS=8$ $\implies \sqrt{61}<8$ i.e. equality does not hold true for any value of $x$ 
Hence there is no solution of the given equation: $6\cos x-5\sin x=8$
A: $6^2 + 5^2 = 61 < 8^2$.
So your attempted answer is ok except for a digit and the fact that you get
$$
\cos(\text{something}) = \frac 8 {\sqrt{61}} >1.
$$
A cosine of a complex number can be bigger than $1$, but a cosine of a real number cannot.
