# How do you solve  sin(x+pi) + cos(x+pi) = 0 from 0 to 2pi?

Feb 2, 2016

$\sin \left(x + \pi\right) = - \sin \left(x\right) = \sin \left(- x\right) = \cos \left(\frac{\pi}{2} + x\right)$

$\cos \left(x + \pi\right) = - \cos \left(x\right)$

So According to the question $\cos \left(\frac{\pi}{2} + x\right) = \cos \left(x\right)$

Using solution of cos we get

$\left(\frac{\pi}{2}\right) + x = 2 n \pi \pm \left(x\right)$

So case I:- $\left(\frac{\pi}{2}\right) + x = 2 n \pi + \left(x\right)$

But here, the value of x is cancelled, so

case II:- $\left(\frac{\pi}{2}\right) + x = 2 n \pi - \left(x\right)$

$x = n \pi - \frac{\pi}{4}$

The solution of this equation is $x = n \pi - \frac{\pi}{4}$