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Yana18102001
@Yana18102001
August 2022
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Решить уравнение
sinx+cosx=1-sin2x
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армения20171
Sinx+cosx=1-sin2x
sinx+cosx=t
(sinx+cosx)²=t²
sin²x+2sinxcosx+cos²x=t²
2sinxcosx=t²-1
sin2x=t²-1
t=1-(t²-1)
t=1-t²+1
t²+t-2=0
D=1+8=9=3²
t=(-1±3)/2
t1=-2;t2=1
1)sinx+cosx=-2;x€∅
2)sinx+cosx=1
√2sin(π/4+x)=1
sin(π/4+x)=√2/2
π/4+x=(-1)ⁿπ/4+πn
x=-π/4+(-1)ⁿπ/4+πn
n=2k
x=2πk
n=2k+1
x=-π/2+π(2k+1)=
π/2+2πk
2 votes
Thanks 1
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Answers & Comments
sinx+cosx=t
(sinx+cosx)²=t²
sin²x+2sinxcosx+cos²x=t²
2sinxcosx=t²-1
sin2x=t²-1
t=1-(t²-1)
t=1-t²+1
t²+t-2=0
D=1+8=9=3²
t=(-1±3)/2
t1=-2;t2=1
1)sinx+cosx=-2;x€∅
2)sinx+cosx=1
√2sin(π/4+x)=1
sin(π/4+x)=√2/2
π/4+x=(-1)ⁿπ/4+πn
x=-π/4+(-1)ⁿπ/4+πn
n=2k
x=2πk
n=2k+1
x=-π/2+π(2k+1)=
π/2+2πk