# Trigonometric equations

1. Sep 1, 2010

### zorro

1. The problem statement, all variables and given/known data

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
2. Relevant equations

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?

2. Sep 1, 2010

### Mentallic

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)=$\pm$1