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Polishaa22

14 нояб. 2014 г., 9:59:58 (9 лет назад)

пусть х/2=α, тогда
sinα+sin3α=√3 cosα
sinα+3sinα-4sin³α-√3cosα=0
4sinα(1-sin²α)-√3cosα=0
4sinα cos²α-√3cosα=0
cosα(4sinαcosα-√3)=0
cosα=0                        4sinαcosα-√3=0
α=π/2 +πn, n∈Z            2sinαcosα=√3/2
x/2=π/2+πn,n∈Z            sin2α=√3 /2
x=π+2πn, n∈Z               2α=π/3 + 2πk, k∈Z
x1=π                            x= π/3 + 2πk, k∈Z
                                    x2=π/3 +2π=7π/3