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lerarhipova189
@lerarhipova189
June 2022
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решите уравнение: sin2x+2cos2x=1
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mikael2
2sinxcosx+2cos²x-2sin²x-cos²x-sin²x=0
2sinxcosx+cos²x-3sin²x=0
поделим на cos²x
2tgx+1-3tg²x=0
3tg²x-2tgx-1=0
D=4+12=16 √16=4
tgx1=1/6[2+4]=1 tgx2=1/6[2-4]=-1/8
x1=π/4+πn x2=arctg(-1/8)+πn n∈Z
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Answers & Comments
2sinxcosx+cos²x-3sin²x=0
поделим на cos²x
2tgx+1-3tg²x=0
3tg²x-2tgx-1=0
D=4+12=16 √16=4
tgx1=1/6[2+4]=1 tgx2=1/6[2-4]=-1/8
x1=π/4+πn x2=arctg(-1/8)+πn n∈Z