What Is the Correct General Solution for the Equation sin(x)sin(y) = 1?

In summary, the general solutions for sinx.siny=1 are x= n(pi)+(pi/2)(-1)^n+1 and y= m(pi)+(pi/2)(-1)^m+1, where n,m belong to I. The first choice is not a correct answer because it does not account for the possibility of sin(x)=-sin(y), which would make the product sin(x)sin(y)=-1. The second and third choices, x=(4n+1)(pi/2) and y=(4m+1)(pi/2) and x=(4n-1)(pi/2) and y=(4m-1)(pi/2),
  • #1
zorro
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Homework Statement



The general solution of sinx.siny=1 is

1.x=n(pi)+(pi/2)(-1)^n+1 y=m(pi)+(pi/2)(-1)^m+1
2.x=(4n+1)(pi/2) y=(4m+1)(pi/2)
3.x=(4n-1)(pi/2) y=(4m-1)(pi/2)

where n,m belong to I

Homework Equations


Answer- 2 and 3

The Attempt at a Solution



This is possible only if sinx=siny=1 or sinx=siny= -1

Substituting arbitrary values of n,m in 1st choice...we get correct answer
but 1st choice is not included in correct answers
WHY?
 
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  • #2
Frankly, I don't see the point of having y=f(n) since y=x.
The first doesn't work because you can choose n and m such that sin(x)=-sin(y) so sin(x)sin(y)=-1

The general solutions are [tex]x= \pi/2 +2\pi n[/tex] such that sin(x)=1, and [tex]y=-\pi /2 +2\pi n[/tex] such that sin(y)=-1. These are equivalent to your second and third equation.

[tex]z=\pi/2+\pi n[/tex] gives sin(z)=[itex]\pm[/itex]1
 

FAQ: What Is the Correct General Solution for the Equation sin(x)sin(y) = 1?

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