What is the general solution for 4cos^2(x)+8cos(x)+4=0 ?

The general solution for 4cos^2(x)+8cos(x)+4=0 is x=pi+2pin

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4cos^2(x)+8cos(x)+4=0

4 cos^{2} (x) + 8 cos (x) + 4 = 0

x = π + 2 πn

Solution steps

4 cos^{2} (x) + 8 cos (x) + 4 = 0

Solve by substitution

cos (x) = - 1

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

x = π + 2 πn

4 sin (x) cos (x) = 22 cos (x)

t, x = cos^{2} (t)

sin (x) = \frac{24}{27}

cot^{2} (x) = 0

4 cos^{2} (x) + 5 sin (x) = 3

What is the general solution for 4cos^2(x)+8cos(x)+4=0 ?
The general solution for 4cos^2(x)+8cos(x)+4=0 is x=pi+2pin