General solution of trigonometric function

**sachinrajsharma** · March 7th 2013, 06:16 AM

For any real numbers x and y. Sinx = siny implies $n\pi + (-1)^ny, where n \in Z$

If sinx = siny, then sinx -siny = 0
or $2cos\frac{x+y}{2}sin\frac{x-y}{2} = 0$

$\Rightarrow cos\frac{x+y}{2} = 0 ; sin\frac{x-y}{2}=0$

Therefore $\frac{x+y}{2} = (2n+1)\frac{\pi}{2} or \frac{x-y}{2} = n\pi,$ where $n \in Z$

i.e. X = $(2n+1)\pi -y or x = 2n\pi +y$ , where $n \in Z$

Please guide further.....

**MINOANMAN** · March 7th 2013, 08:37 AM

since you have found (x+y)/2 and (x-y)/2 make two similtaneous equations and solve them for x ,y it is easy .

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