Question

Find the values of θ satisfying sin θ > 0.

Answer 1

sinθ is positive in quadrants 1 and 2. So 0 < θ < π. But quadrants 1 and 2 can also be written as 2π < θ < 3π or 4π < θ < 5π. So, the general solution can be obtained by adding 2nπ to the first solution. That is,

2nπ < θ < (2n + 1) πn∈I