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Уникум007
@Уникум007
August 2022
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решите уравнение
2sin²x+sinx*cosx-3cos²x=0
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DChernoivanova
2sin²x+sin x×cos x-3cos²x=0 |:cos²x; cos²x≠0, cos x≠0
2tg²x+tg x-3=0
Пусть tg x=a, тогда
2a²+a-3=0
D=1-4×2×(-3)=1+24=25=5²
a₁=(-1+5)/2×2=4/4=1
a₂=(-1-5)/2×2=-6/4=-2/3
tg x=1
x=π/4+πn, n∈Z
tg x=-2/3
x=arctg(-2/3)+πm, m∈Z
Ответ: π/4+πn, n∈Z; arctg(-2/3)+πm, m∈Z
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Answers & Comments
2tg²x+tg x-3=0
Пусть tg x=a, тогда
2a²+a-3=0
D=1-4×2×(-3)=1+24=25=5²
a₁=(-1+5)/2×2=4/4=1
a₂=(-1-5)/2×2=-6/4=-2/3
tg x=1
x=π/4+πn, n∈Z
tg x=-2/3
x=arctg(-2/3)+πm, m∈Z
Ответ: π/4+πn, n∈Z; arctg(-2/3)+πm, m∈Z