Ask Me Help Desk - Trig Equations

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:

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

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

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