<s>[INST] 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:
$$
 -\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z
$$
 [/INST] $$6\cos { x } -5\sin { x } =8\\ 6\cos ^{ 2 }{ \frac { x }{ 2 } -6\sin ^{ 2 }{ \frac { x }{ 2 } -10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } =8 } \cos ^{ 2 }{ \frac { x }{ 2 }  } +8\sin ^{ 2 }{ \frac { x }{ 2 }  }  }  }  } \\ 14\sin ^{ 2 }{ \frac { x }{ 2 }  } +10\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 }  }  } +2\cos ^{ 2 }{ \frac { x }{ 2 }  } =0\\ \cos ^{ 2 }{ \frac { x }{ 2 }  } \neq 0$$
$$14\tan ^{ 2 }{ \frac { x }{ 2 } +10\tan { \frac { x }{ 2 } +2=0 }  } \\ \tan { \frac { x }{ 2 } =a } \\ 7{ a }^{ 2 }+5a+1=0\\ $$
can you take from here?
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