4cos(2x-pi/2)

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Solution for 4cos(2x-pi/2) equation: 


x in (-oo:+oo)

4*cos(2*x-(pi/2)) = 0

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

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

t_1 = pi*k_1+pi/2

2*x-t_1-(pi/2) = 0

2*x-t_1-(pi/2) = 0 // + -t_1-(pi/2)

2*x = -(-t_1-(pi/2)) // : 2

x = (-(-t_1-(pi/2)))/2

x = (t_1+pi/2)/2

x = (pi*k_1+pi/2+pi/2)/2 i k_1 należy do I

4*cos(2*x-(pi/2)) = 0  <=>  4 = 0 or 4*cos(2*x-(pi/2)) = 0  <=>  cos(2*x-(pi/2)) = 0

x = (pi*k_1+pi/2+pi/2)/2

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