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maly6ka
@maly6ka
July 2022
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Тригонометрические уравнения
sin2x+4cos^2x=1
tg^2x - 3/ctgx-4=0
4sin^2x-sin2x=0
sin(x/3+П/4)= -√3/2
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lubavka13
2) tg^2x -3tgx-4=0D=25tgx =4, x=arctg4+πk, k€Z.tgx=-1, x=¾π+πn, n€Z.
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lubavka13
Во втором 2sinx выносится за скобки (синус двойного угла)
lubavka13
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