<s>[INST] Determine the general solution for $2\cos^2x-5\cos x+2=0$ Determine the general solution for $ 2\cos^{2}x-5\cos x+2=0$
My attempt: 
$2u^2 - 5u + 2 = 0$
$(2u - 1)(u - 2) = 0$
$u = \frac{1}{2}$ or $u = 2$
$\cos(x) = \frac{1}{2}$ or $\cos(x) = 2$
The answers on the worksheet are as follows but my answer does not match...
A. $2.89$, $3.39$
B. $3.39$, $6.03$
C. $0.25$, $6.03$
D. $4.46$, $6.03$
 [/INST] [\begin{array}{l}
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{\rm{                }}2{\cos ^2}x - 5\cos x + 2 = 0\\
2\cos x(\cos x - 2) - 1(\cos x - 2) = 0\\
{\rm{               }}(2\cos x - 1)(cosx - 2) = 0\\
2\cos x - 1 = 0{\rm{   OR    }}\cos x - 2 = 0\\
\\
{\rm{Case 1}}\\
2\cos x - 1 = 0\\
{\rm{       }}\cos x = \frac{1}{2}\\
{\rm{      }}\cos x{\rm{ }} = \cos \frac{\pi }{3}\\
{\rm{         }}\therefore x  =  2n\pi  \pm \frac{\pi }{3}{\rm{      }}\left( {n \in Z} \right)\\
\\
{\rm{Case2}}\\
\cos x - 2 = 0\\
{\rm{       }}\cos x = 2\\
{\rm{but }}\\
{\rm{       cosx}} \ne 2\\
{\rm{because    - 1}} \le {\rm{cosx}} \le {\rm{1 }}\\
\therefore {\rm{   }}In{\rm{ }}this{\rm{ }}case{\rm{ }},there{\rm{ }}is{\rm{ }}no{\rm{ }}solution{\rm{ }}for{\rm{ }}\cos x{\rm{  }}
\end{array}]
</s>