what is the interval you seek solutions in? Because in every interval \left(\frac{k \pi}{2}, \frac{(k+2)\pi}{2} \right) with k an odd integer, you will find solutions.
1 Answers
Hari Shankar
·2009-09-18 20:51:23
what is the interval you seek solutions in? Because in every interval \left(\frac{k \pi}{2}, \frac{(k+2)\pi}{2} \right) with k an odd integer, you will find solutions.