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bezalloff
@bezalloff
August 2022
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РОЗВЯЖІТЬ РІВНЯННЯ, БУДЬ ЛАСОЧКА!
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Armenia2780
Sin²x+2sinxcosx-3cos²x=0 /cos²x≠0
tg²x+2tgx-3=0
tgx=t
t²+2t-3=0
D=4+12=16=4²
t=(-2±4)/2
t1=-3
t2=1
1)tgx=-3
x=-arctg3+πk;k€Z
2)tgx=1
x=π/4±πk;k€Z
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Answers & Comments
tg²x+2tgx-3=0
tgx=t
t²+2t-3=0
D=4+12=16=4²
t=(-2±4)/2
t1=-3
t2=1
1)tgx=-3
x=-arctg3+πk;k€Z
2)tgx=1
x=π/4±πk;k€Z