Find the solution set of each equation if 0° ≤ θ < 360°. (Please explain it with proper solutions)

Question

(sin θ)^2 + 2 sin θ + 1 = 0

Answer 1

We can solve sin² θ + 2 sin θ + 1 = 0 because it is quadratic in nature. If we visualize sin θ being replaces with x, then we simply have x² + 2x + 1 = 0. So we will solve our problem the same way as this quadratic - by factoring.

We can factor the left side of our equation into:

(sin θ + 1)(sin θ + 1) = 0 (you can verify by distributing, but this should be confirmed just with basic factoring skills)

Now we solve by setting each factor equal to 0. Since both factors are identical, we only have to do this for one.

sin θ + 1 = 0

sin θ = -1

and using the unit circle, we know the only location where sin θ = -1 is at 270^o.

so θ = 270^o