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millar2009S · May 20, 2009, 10:50 AM

2sin2x+1 =0 :confused:

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:

$2sin(2x)+1=0$

$sin(2x)=\frac{-1}{2}$

$x=\frac{sin^{-1}(\frac{-1}{2})}{2}$