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Dreymers
@Dreymers
August 2022
1
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sinx+2=0
sin2x=-√3/2
2cosx-√2=0
cos5x=0
√3tg2x+1=0
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acar
1) sinx=-2
x=(-1)ⁿarcsin(-2)+πn, n∈Z
2) sin2x=-√3/2
x=(-1)ⁿ arcsin(-√3/2)+πn , n∈Z
x=-( (-1)ⁿ *π/3)+πn , n∈Z
3)2cosx=√2
cosx=√2/2
x=+-arccos(√2/2)+2πn , n∈Z
x=+-π/4+2πn , n∈Z
4) cos5x=0
5x=π/2+πn , n∈Z
x=π/10+πn/5 , n∈Z
5) tg2x=-1/√3
x=-arctg(1/√3)+πn , n∈Z
2x=-π/6++πn , n∈Z
x=-π/12+πn/2 , n∈Z
1 votes
Thanks 1
Dreymers
Извините,но вы не могли бы обьяснить как решать 2-е?
acar
для каждого случая есть формула , можешь в интернете посмотреть sinx=a
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Answers & Comments
x=(-1)ⁿarcsin(-2)+πn, n∈Z
2) sin2x=-√3/2
x=(-1)ⁿ arcsin(-√3/2)+πn , n∈Z
x=-( (-1)ⁿ *π/3)+πn , n∈Z
3)2cosx=√2
cosx=√2/2
x=+-arccos(√2/2)+2πn , n∈Z
x=+-π/4+2πn , n∈Z
4) cos5x=0
5x=π/2+πn , n∈Z
x=π/10+πn/5 , n∈Z
5) tg2x=-1/√3
x=-arctg(1/√3)+πn , n∈Z
2x=-π/6++πn , n∈Z
x=-π/12+πn/2 , n∈Z