(2cos(2x))/2

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Solution for (2cos(2x))/2 equation:


x in (-oo:+oo)

(2*cos(2*x))/2 = 0

(2*cos(2*x))/2 = 0 // * 2

2*cos(2*x) = 0

cos(2*x) = 0

cos(2*x) = 0 <=> 2*x = pi*k_1+pi/2 i k_1 należy do I

t_1 = pi*k_1+pi/2

2*x-t_1 = 0

2*x-t_1 = 0 // + t_1

2*x = t_1 // : 2

x = t_1/2

x = pi*k_1+pi/2/2 i k_1 należy do I

2*cos(2*x) = 0 <=> 2 = 0 or 2*cos(2*x) = 0 <=> cos(2*x) = 0

x = pi*k_1+pi/2/2

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