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 ≤ 2

θ< 4π.

The reference angle is

`α = arcsin 0.8 = 0.9273`

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

2

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

That is

`2θ = 0.9273, 2.2143, 7.2105, 8.4975`

But we need values for *θ*, not 2*θ*,
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.