galactus
May 20, 2009, 11:10 AM
Since trig functions are periodic, they generally have several solutions over a given interval.
2sin(2x)+1=0
With that, when we solve this one we have:
x=C{\pi}+\frac{7\pi}{12}, \;\ x=C{\pi}-\frac{\pi}{12}
Let C=0, 1, and so forth. I assume your interval is [0,2\pi]
Find all solutions over this interval should suffice.
To find something to start with:
2sin(2x)+1=0
sin(2x)=\frac{-1}{2}
x=\frac{sin^{-1}(\frac{-1}{2})}{2}