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Malumtt
@Malumtt
December 2019
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Obtenha m, m E IR, de modo que:
sen x = m/5 e cos x = m + 1/5
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lucasblack
Sen*2x+ cós*2x=1
(m/5)*2+(m+1)*2/5*2=1
m*2/5*2+(m*2+2m+1)/25=1
m*2/25+(m*2+2m+1)/25=1
m*2+m*2+2m+1=25•1
2m*2+2m=25-1
2m*2+2m-24=0 ÷2
m*2+m-12=0
fórmula de bascara
x=(-1+-√1-4•1•(-12))/2
x=(-1+-√1+48)/2
x=(-1+-√49)/2
x(-1+-7)/2
x1=-1+7/2=6/2=3
x2=-1-7/2=-8/2=-4
logo para m{-4,3}
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(m/5)*2+(m+1)*2/5*2=1
m*2/5*2+(m*2+2m+1)/25=1
m*2/25+(m*2+2m+1)/25=1
m*2+m*2+2m+1=25•1
2m*2+2m=25-1
2m*2+2m-24=0 ÷2
m*2+m-12=0
fórmula de bascara
x=(-1+-√1-4•1•(-12))/2
x=(-1+-√1+48)/2
x=(-1+-√49)/2
x(-1+-7)/2
x1=-1+7/2=6/2=3
x2=-1-7/2=-8/2=-4
logo para m{-4,3}