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

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SUMMARY

The correct general solutions for the equation sin(x)sin(y) = 1 are given by the equations x = (4n + 1)(π/2) and y = (4m + 1)(π/2), as well as x = (4n - 1)(π/2) and y = (4m - 1)(π/2), where n and m are integers. The first proposed solution, x = n(π) + (π/2)(-1)^n + 1 and y = m(π) + (π/2)(-1)^m + 1, is invalid because it allows for the possibility of sin(x) = -sin(y), resulting in sin(x)sin(y) = -1. The valid solutions ensure that sin(x) and sin(y) are both equal to 1 or -1, confirming the correctness of the second and third equations.

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

z=\pi/2+\pi n gives sin(z)=\pm1
 

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