solve for values of x for 0<= x < 2pi
cos^2 x +2 cos x +1 = 0
we have : cos^2 x +2 cos x +1 = 0
we know : (x+1)²=x²+2x+1
so : cos^2 x +2 cos x +1 = (cos(x) + 1)² = 0
so : cos(x) + 1 =0
so : cos(x) = -1
so : cos(x) = cos(pi)
so : x= pi + 2Kpi
or : x= - pi + 2Kpi
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