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Anthony301997
@Anthony301997
July 2022
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Ребят, очень вас прошу, решите пожалуйста, что кружком отмечено:), хотя бы несколько, буду очень рад:)
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nikolson93
В) √3 * tg²3x - 3*tg3x=0
tg3x=a
√3*a² - 3a=0
a(√3*a-3)=0
a=0 a=√3
tg3x=0 tg3x=√3
3x=πт 3x=π/3+πn
г) 4sin²(2x+π/3) =1
sin²(2x+π/3)=1/4
sin(2x+π/3) = 1/2 sin(2x+π/3)= -1/2
2x+π/3 =(-1)^n * π/6+πn x=(-1)^(n+1)*π/12 -π/6 + πn/2
x=(-1)^n*π/12 - π/6 + πn/2
a) sin2x=cos2x
tg2x=1
2x=π/4+πn
x=π/8 + πn/2
1 votes
Thanks 1
Anthony301997
спасибо;)
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Answers & Comments
tg3x=a
√3*a² - 3a=0
a(√3*a-3)=0
a=0 a=√3
tg3x=0 tg3x=√3
3x=πт 3x=π/3+πn
г) 4sin²(2x+π/3) =1
sin²(2x+π/3)=1/4
sin(2x+π/3) = 1/2 sin(2x+π/3)= -1/2
2x+π/3 =(-1)^n * π/6+πn x=(-1)^(n+1)*π/12 -π/6 + πn/2
x=(-1)^n*π/12 - π/6 + πn/2
a) sin2x=cos2x
tg2x=1
2x=π/4+πn
x=π/8 + πn/2