Answered

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solve 2cos^2x+5cosx+2=0

Sagot :

marmal
[tex]2cos^2x+5cosx+2=0\\ x\in<-1;1>\\ \Delta=b^2-4ac\\ \Delta=9\\ \sqrt{\Delta}=3\\ cosx=1\\ or\\ cosx=-6\\ -6\notin<-1;1>\\[/tex]

Then using trygonometric table or a graph we read that:
[tex]cosx=1\Leftrightarrow(x)=0[/tex]

Period of cosinus function = 2π
So the last answer is [tex]x=0+2k\pi\\[/tex]
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