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D( x )
sin(2*x) = 0
sin(2*x) = 0
sin(2*x) = 0
sin(2*x) = 0 <=> 2*x = pi*k_1 i k_1 należy do I
t_1 = pi*k_1
2*x-t_1 = 0
2*x-t_1 = 0 // + t_1
2*x = t_1 // : 2
x = t_1/2
x = pi*k_1/2 i k_1 należy do I
x in {( -oo : +oo ) / < pi*k_1/2 : pi*k_1/2 >} i k_1 -> {I}
sin(5*x)/sin(2*x) = 0
sin(5*x)/sin(2*x) = 0 // * sin(2*x)
sin(5*x) = 0
sin(5*x) = 0 <=> 5*x = pi*k_1 i k_1 należy do I
t_1 = pi*k_1
5*x-t_1 = 0
5*x-t_1 = 0 // + t_1
5*x = t_1 // : 5
x = t_1/5
x = pi*k_1/5 i k_1 należy do I
x = pi*k_1/5
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