Let's determine the solutions of the following:
[tex]\text{ 2 }\cdot\text{ Sin \lparen}\theta)\text{ = 1}[/tex]
We get,
[tex]\text{ 2 }\cdot\text{ Sin \lparen}\theta)\text{ = 1}[/tex][tex]\text{ Sin \lparen}\theta)\text{ = }\frac{1}{2}[/tex][tex]\text{ }\theta\text{ = Sin}^{-1}(\frac{1}{2})[/tex][tex]\text{ }\theta\text{ = }\frac{\pi}{6}[/tex]
Since we are asked for the solution at 0 < Θ < 2π interval, we will be looking at the solution in the 1st and second quadrants.
For the second solution, subtract the reference angle from π to find the solution in the second quadrant.
We get,
[tex]\text{ }\pi\text{ - }\frac{\pi}{6}\text{ = }\frac{6\pi}{6}\text{ - }\frac{\pi}{6}[/tex][tex]\text{ = }\frac{5\pi}{6}[/tex]
Therefore, the solutions are:
[tex]\text{ }\frac{\pi}{6},\text{ }\frac{5\pi}{6}[/tex]
