решите ур-ие cos2x-3cosx+2=0

[tex]cos2x-3cosx+2=0; \\ 2cos^{2}x-1-3cosx+2=0; \\ 2cos^{2}x-3cosx+1=0; \\ cosx=t; \\ 2t^{2}-3t+1=0; \\ D=9-8=1; \\ t_{1} = \frac{3+1}{4}=1; t_{2} = \frac{3-1}{4}= \frac{1}{2} ; \\ 1) cosx=1; x=2 \pi n \\ 2)cosx= \frac{1}{2} ; x=+- acrcos \frac{1}{2} +2 \pi k.; x=+- \frac{ \pi }{3} +2 \pi k[/tex]

n∈Z
k∈Z