2sin2x+1 =0 :confused:

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}