e^(xy) + y^2 = sin(x + y) 

asked Dec 9, 2016 in Algebra 2 Answers by Mathical Level 10 User (55,380 points)

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1 Answer

It's unlikely that y can be expressed in terms of x but we can make some observations.

The expression on the right has a range of -1 to 1, but the expression on the left is always strictly positive. Therefore we know that 0<sin(x+y)≤1 and 0<e^(xy)+y^2≤1. 

When y=0, x=(4n+1)π/2 where n is an integer, because sin(x)=1 when y=0.

So for one value of y there are multiple values of x.

When x=0, y^2=sin(y)-1, no solutions for y. When x<0 and y>0, y≤1-e^(xy) so y<1 (radians).

When e^(xy)<1, xy<0 and y>0, so 0≤y<1 and x<0 (or -1<y≤0 and x>0).

answered Dec 18, 2016 by Rod Top Rated User (424,700 points)
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