2cos(1/2x^2)

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Solution for 2cos(1/2x^2) equation: 


x in (-oo:+oo)

2*cos((1/2)*x^2) = 0

cos((1/2)*x^2) = 0

cos((1/2)*x^2) = 0  <=>  (1/2)*x^2 = pi*k_1+pi/2 i k_1 należy do I

t_1 = pi*k_1+pi/2

(1/2)*x^2-t_1 = 0

1/2*x^2 = t_1 // : 1/2

x^2 = 2*t_1

x^2 = 2*t_1 // ^ 1/2

abs(x) = 2^(1/2)*t_1^(1/2)

x = 2^(1/2)*t_1^(1/2) or x = -(2^(1/2)*t_1^(1/2))

x = 2^(1/2)*pi*k_1+pi/2^(1/2) i k_1 należy do I

x = -(2^(1/2)*pi*k_1+pi/2^(1/2)) i k_1 należy do I

2*cos((1/2)*x^2) = 0  <=>  2 = 0 or 2*cos((1/2)*x^2) = 0  <=>  cos((1/2)*x^2) = 0

x in { 2^(1/2)*pi*k_1+pi/2^(1/2), -(2^(1/2)*pi*k_1+pi/2^(1/2)) }

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