If the problem involved θ only, we would expect 2 solutions; one in the first quadrant and one in the second quadrant.

But here our problem involves 2θ, so we have to double the domain (θ values) to account for all possible solutions.

We proceed as follows:

We solve

sin 2θ = 0.8 for 0 ≤ 2θ < 4π.

The reference angle is

α = arcsin 0.8 = 0.9273

So the values for 2θ will be in quadrants I, II, V, VI.

2θ = 0.9273, π − 0.9273, 2π + 0.9273, 3π− 0.9273

That is

2θ = 0.9273, 2.2143, 7.2105, 8.4975

But we need values for θ, not 2θ, so we divide throughout by 2:

θ = 0.4637, 1.1072, 3.6053, 4.2488

Are our answers correct? As usual, we will check by graphing the original expression:

We can see from the graph that our 4 values are reasonable, since these are the only 4 values that satisfy sin 2θ = 0.8.

Note: We can always check our solutions with calculator, but it is easy to "miss out" on some of the required values if we only use calculator.