Answer:
[tex]\displaystyle \theta = \left\{\frac{7\pi}{6}, \, \frac{3\pi}{2}, \frac{11\pi}{6}\right\}[/tex]
Step-by-step explanation:
We want to solve the equation:
[tex]2\sin ^2 \theta + 3\sin \theta + 1 = 0[/tex]
In the interval [0, 2π).
Notice that this is in quadratic form. Namely, by letting u = sin(θ), we acquire:
[tex]\displaystyle 2u^2 + 3u + 1 = 0[/tex]
Factor:
[tex]\displaystyle (2u+1)(u+1) = 0[/tex]
By the Zero Product Property:
[tex]\displaystyle 2u + 1 = 0\text{ or } u + 1 = 0[/tex]
Solve for each case:
[tex]\displaystyle u = -\frac{1}{2} \text{ or } u = -1[/tex]
Back-substitute:
[tex]\displaystyle \sin \theta = -\frac{1}{2} \text{ or } \sin \theta =-1[/tex]
Use the Unit Circle. Hence, our three solutions in the interval [0, 2π) are:
[tex]\displaystyle \theta = \left\{\frac{7\pi}{6}, \, \frac{3\pi}{2}, \frac{11\pi}{6}\right\}[/tex]
