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:
`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, or π − 0.9273, or 2π + 0.9273, or 3π − 0.9273
`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.