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

Good Luck my Friends

and

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From your friend

Karim

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