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lesnyxalex
@lesnyxalex
July 2022
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3^sin^2 х+3^cos^2х=4 как решить
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армения20171
3^sin^2x+3^(1-sin^2x)=4
3^sin^2x+3/3^sin^2x=0
3^sin^4x+3-4•3^sin^2=0
3^sin^2x=t
t^2+3-4t=0
t^2-4t+3=0
D=16-12=4>0
t1=(4+2)/2=3
t2=(4-2)/2=1
1)3^sin^2x=3
sin^2x=1
sinx=1 x=π/2+2πn
sinx=-1 x1=-π/2+2πk
2)3^sin^2x=1=3^0
sin^2x=0
sin2x=0
2x=2πn
x2=πn n€Z
ответ π/2+2πn;-π/2+2πk;πn;n€Z;k€Z
думаю всё
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lesnyxalex
Спасибо тебе)))
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Answers & Comments
3^sin^2x+3/3^sin^2x=0
3^sin^4x+3-4•3^sin^2=0
3^sin^2x=t
t^2+3-4t=0
t^2-4t+3=0
D=16-12=4>0
t1=(4+2)/2=3
t2=(4-2)/2=1
1)3^sin^2x=3
sin^2x=1
sinx=1 x=π/2+2πn
sinx=-1 x1=-π/2+2πk
2)3^sin^2x=1=3^0
sin^2x=0
sin2x=0
2x=2πn
x2=πn n€Z
ответ π/2+2πn;-π/2+2πk;πn;n€Z;k€Z
думаю всё