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Panny99
@Panny99
July 2022
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Помогите, пожалуйста!!!
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sedinalana
Verified answer
ОДЗ
сosx>0⇒x∈(-π/2+2πn;π/2+2πn)
log(3)(2cosx)=a
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2⇒log(3)(2cosx)=1/2⇒2cosx=√3⇒cosx=√3/2⇒x=+-π/6+2πn
x=5π/3,x=7π/3
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Answers & Comments
Verified answer
ОДЗсosx>0⇒x∈(-π/2+2πn;π/2+2πn)
log(3)(2cosx)=a
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2⇒log(3)(2cosx)=1/2⇒2cosx=√3⇒cosx=√3/2⇒x=+-π/6+2πn
x=5π/3,x=7π/3