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August 2022
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NeZeRAvix
Verified answer
ОДЗ: x+π/6>0 ⇒ x>-π/6
С учетом ОДЗ:
Ответ:
2 votes
Thanks 1
5ViSkaS5
Спасибо <З
армения20171
{2sin²x-5sinx-3=0
{x+π/6>0
1)2sin²x-5sinx-3=0
sinx=t
2t²-5t-3=0
t=(5±7)/4
t1=3
t2=-1/2
sinx=3;x€∅
sinx=-1/2
x=(-1)ⁿ(-π/6)+πn
1)n чётний
x1=-π/6+2πk
2)n не чётний
x2=7π/6+2πn
2)x1+π/6>0
-π/6+2πk+π/6>0
2πk>0
k>0
x2+π/6>0
7π/6+2πn+π/6>0
8π/6+2πn>0
8π+12πn>0
n>-8π/12π
n>-2/3
n={0;1;2;3;...........}
[k={1;2;3;........};x1=-π/6+2πk
[n={0;1;2;3;. ...};x2=7π/6+2πn
1 votes
Thanks 1
5ViSkaS5
Спасибо ~~~
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Answers & Comments
Verified answer
ОДЗ: x+π/6>0 ⇒ x>-π/6
С учетом ОДЗ:
Ответ:
{x+π/6>0
1)2sin²x-5sinx-3=0
sinx=t
2t²-5t-3=0
t=(5±7)/4
t1=3
t2=-1/2
sinx=3;x€∅
sinx=-1/2
x=(-1)ⁿ(-π/6)+πn
1)n чётний
x1=-π/6+2πk
2)n не чётний
x2=7π/6+2πn
2)x1+π/6>0
-π/6+2πk+π/6>0
2πk>0
k>0
x2+π/6>0
7π/6+2πn+π/6>0
8π/6+2πn>0
8π+12πn>0
n>-8π/12π
n>-2/3
n={0;1;2;3;...........}
[k={1;2;3;........};x1=-π/6+2πk
[n={0;1;2;3;. ...};x2=7π/6+2πn