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nicesmall
@nicesmall
July 2022
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Срочно нужна помощь!!!
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amin07am
Verified answer
Ответ ответ ответ ответ ответ ответ ответ
2 votes
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nicesmall
Ты лучший, чел
amin07am
Спасибо@
sedinalana
Verified answer
1
(1-cosx)/sinx/2=2
2sin²x/2/sinx/2=2
2sinx/2=2
sinx/2=1
x/2=π/2+2πn
x=π+4πn,n∈z
2
4ctgx/(1+ctg²x)+sin²2x+1=0
4cosx/sinx:1/sin²x +sin²2x+1=0
4cosxsin²x/sinx+sin²2x+1=0
4cosxsinx+sin²2x+1=0
2sin2x+sin²2x=1=0
(sin2x+1)²=0
sin2x+1=0
sin2x=-1
2x=-π/2+2πn
x=-π/4+πn,n∈z
3
sin6x+sin2x=1/2*tg2x
2sin4xcos2x=1/2*sin2x/cos2x
4sin4xcos²2x=sin2x
8sin2xcos³2x-sin2x=0
sin2x*(8cos³2x-1)=0
sin2x=0⇒2x=πn⇒x=πn/2,n∈z
8cos³2x-1=0⇒8cos³2x=1⇒cos³2x=1/8⇒cos2x=1/2⇒2x=+-π/3+2πk⇒
x=+-π/6+πk,k∈z
0 votes
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Verified answer
Ответ ответ ответ ответ ответ ответ ответVerified answer
1(1-cosx)/sinx/2=2
2sin²x/2/sinx/2=2
2sinx/2=2
sinx/2=1
x/2=π/2+2πn
x=π+4πn,n∈z
2
4ctgx/(1+ctg²x)+sin²2x+1=0
4cosx/sinx:1/sin²x +sin²2x+1=0
4cosxsin²x/sinx+sin²2x+1=0
4cosxsinx+sin²2x+1=0
2sin2x+sin²2x=1=0
(sin2x+1)²=0
sin2x+1=0
sin2x=-1
2x=-π/2+2πn
x=-π/4+πn,n∈z
3
sin6x+sin2x=1/2*tg2x
2sin4xcos2x=1/2*sin2x/cos2x
4sin4xcos²2x=sin2x
8sin2xcos³2x-sin2x=0
sin2x*(8cos³2x-1)=0
sin2x=0⇒2x=πn⇒x=πn/2,n∈z
8cos³2x-1=0⇒8cos³2x=1⇒cos³2x=1/8⇒cos2x=1/2⇒2x=+-π/3+2πk⇒
x=+-π/6+πk,k∈z