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Solución de sistemas de ecuaciones/ sinx+siny=1; x+y=p/3

sinx+siny=1; x+y=p/3

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Solución

Ha introducido [src] 
sin(x) + sin(y) = 1
sin⁡(x)+sin⁡(y)=1\sin{\left(x \right)} + \sin{\left(y \right)} = 1 
        p
x + y = -
        3
x+y=p3x + y = \frac{p}{3} 
x + y = p/3
Respuesta rápida
p1=3y−3asin⁡(sin⁡(y)−1)p_{1} = 3 y - 3 \operatorname{asin}{\left(\sin{\left(y \right)} - 1 \right)}
=
3y−3asin⁡(sin⁡(y)−1)3 y - 3 \operatorname{asin}{\left(\sin{\left(y \right)} - 1 \right)}
=
3*y - 3*asin(-1 + sin(y))

x1=−asin⁡(sin⁡(y)−1)x_{1} = - \operatorname{asin}{\left(\sin{\left(y \right)} - 1 \right)}
=
−asin⁡(sin⁡(y)−1)- \operatorname{asin}{\left(\sin{\left(y \right)} - 1 \right)}
=
-asin(-1 + sin(y))
p2=3y+3asin⁡(sin⁡(y)−1)+3πp_{2} = 3 y + 3 \operatorname{asin}{\left(\sin{\left(y \right)} - 1 \right)} + 3 \pi
=
3y+3asin⁡(sin⁡(y)−1)+3π3 y + 3 \operatorname{asin}{\left(\sin{\left(y \right)} - 1 \right)} + 3 \pi
=
9.42477796076938 + 3*y + 3*asin(-1 + sin(y))

x2=asin⁡(sin⁡(y)−1)+πx_{2} = \operatorname{asin}{\left(\sin{\left(y \right)} - 1 \right)} + \pi
=
asin⁡(sin⁡(y)−1)+π\operatorname{asin}{\left(\sin{\left(y \right)} - 1 \right)} + \pi
=
3.14159265358979 + asin(-1 + sin(y))
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