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Sveta123455
@Sveta123455
May 2022
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sin2x+5(cosx+sinx+1)=0
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армения20171
Verified answer
Sin2x+5(cosx+sinx+1)=0
sin2x+1+5(cosx+sinx)+4=0
1)(2sinxcosx+sin²x+cos²a)+5(cosx+sinx)+4=0
(cosx+sinx)²+5(cosx+sinx)+4=0
cosx+sinx=t
t²+5t+4=0
D=25-16=9=3²
t=(-5±3)/2
t1=-4;t2=-1
2) sinx+cosx=√2(1/√2sinx+1/√2cosx)=
√2sin(x+π/4)
t€[-√2;√2]
t1=-4¢[-√2;√2]
3)√2sin(x+π/4)=-1
sin(x+π/4)=-1/√2
x+π/4=(-1)ⁿ(-π/4)+πn
x=-π/4+(-1)ⁿ(-π/4)+πn;n€Z
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Answers & Comments
Verified answer
Sin2x+5(cosx+sinx+1)=0sin2x+1+5(cosx+sinx)+4=0
1)(2sinxcosx+sin²x+cos²a)+5(cosx+sinx)+4=0
(cosx+sinx)²+5(cosx+sinx)+4=0
cosx+sinx=t
t²+5t+4=0
D=25-16=9=3²
t=(-5±3)/2
t1=-4;t2=-1
2) sinx+cosx=√2(1/√2sinx+1/√2cosx)=
√2sin(x+π/4)
t€[-√2;√2]
t1=-4¢[-√2;√2]
3)√2sin(x+π/4)=-1
sin(x+π/4)=-1/√2
x+π/4=(-1)ⁿ(-π/4)+πn
x=-π/4+(-1)ⁿ(-π/4)+πn;n€Z