2cos(5x/2)cos(-x/2)=0

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Solution for 2cos(5x/2)cos(-x/2)=0 equation: 


x in (-oo:+oo)

2*cos((5*x)/2)*cos(-x/2) = 0

2*cos((5*x)/2) = 0

cos((5*x)/2) = 0

cos((5*x)/2) = 0  <=>  (5*x)/2 = pi*k_1+pi/2 i k_1 należy do I

t_1 = pi*k_1+pi/2

(5*x)/2-t_1 = 0

5/2*x-t_1 = 0 // + t_1

5/2*x = t_1 // : 5/2

x = t_1/5/2

x = 2/5*t_1

x = 2/5*pi*k_1+pi/2 i k_1 należy do I

2*cos((5*x)/2) = 0  <=>  2 = 0 or 2*cos((5*x)/2) = 0  <=>  cos((5*x)/2) = 0

cos(-x/2) = 0

cos(-x/2) = 0  <=>  -x/2 = pi*k_2+pi/2 i k_2 należy do I

t_2 = pi*k_2+pi/2

-t_2-x/2 = 0

-t_2-1/2*x = 0 // + t_2

-1/2*x = t_2 // : -1/2

x = t_2/(-1/2)

x = -2*t_2

x = -2*pi*k_2+pi/2 i k_2 należy do I

2*cos((5*x)/2)*cos(-x/2) = 0  <=>  2*cos((5*x)/2) = 0 or 2*cos((5*x)/2)*cos(-x/2) = 0  <=>  cos(-x/2) = 0

x in { 2/5*pi*k_1+pi/2, -2*pi*k_2+pi/2 }

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