Originally Posted by

**mr fantastic** k is an integer. The general solution of sin(A) = 0 is A = k pi, that is, $\displaystyle \pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$

Similarly for cos(A) = 0: $\displaystyle A = \frac{(2k + 1)\pi}{2}$, that is, $\displaystyle \pm \frac{\pi}{2}, \, \pm \frac{3\pi}{2} \, \pm \frac{5\pi}{2} .......$. Peritus made a small mistake here by the way - it's NOT $\displaystyle \frac{\pi}{2}k$ since this also gives $\displaystyle \pm \pi, \, \pm 2\pi, \, \pm 3 \pi .......$ and obviously cos is NOT 0 for those values ....