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deanonm
@deanonm
July 2022
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2log^2 3(2cosx)-7log3(2cosx)+3=0
на отрезке [-3pi;-3pi/2]
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sangers1959
Verified answer
2*log²₃(2*cosx)-7*log₃(2*cosx)+3=0 [-3π;-3π/2]
ОДЗ: 2*cosx>0 cosx>0 x∈(-π/2;π/2)
log₃(2*cosx)=t
2t²-7t+3=0 D=25
t₁=3 ⇒ 2*cosx=3 cosx=1,5 ∉ так как |cosx|≤1
t₂=0,5 ⇒ 2*cosx=1/2 cosx=1/4
x₁,₂=+/-arccos(1/4).
1 votes
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deanonm
при обратной замене log забыли
sangers1959
t₁=3 ⇒ log₃(2*cosx)=3 2*cosx=3^3=27 cosx=13,5 ∉ так как |cosx|≤1
t₂=0,5 ⇒ log₃(2*cosx)=1/2 2*cosx=3^(1/2) cosx=3^(1/2)/2 x₁=π/3+2πn x₂=-π/3+2πn.
deanonm
x=+- Пи/6 + 2 Пиn
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Answers & Comments
Verified answer
2*log²₃(2*cosx)-7*log₃(2*cosx)+3=0 [-3π;-3π/2]ОДЗ: 2*cosx>0 cosx>0 x∈(-π/2;π/2)
log₃(2*cosx)=t
2t²-7t+3=0 D=25
t₁=3 ⇒ 2*cosx=3 cosx=1,5 ∉ так как |cosx|≤1
t₂=0,5 ⇒ 2*cosx=1/2 cosx=1/4 x₁,₂=+/-arccos(1/4).
t₂=0,5 ⇒ log₃(2*cosx)=1/2 2*cosx=3^(1/2) cosx=3^(1/2)/2 x₁=π/3+2πn x₂=-π/3+2πn.