Question

Can someone show me how to factor this please?

Answer 1

Answer:

$\displaystyle \theta = \left\{\frac{7\pi}{6}, \, \frac{3\pi}{2}, \frac{11\pi}{6}\right\}$

Step-by-step explanation:

We want to solve the equation:

$2\sin ^2 \theta + 3\sin \theta + 1 = 0$

In the interval [0, 2π).

Notice that this is in quadratic form. Namely, by letting u = sin(θ), we acquire:

$\displaystyle 2u^2 + 3u + 1 = 0$

Factor:

$\displaystyle (2u+1)(u+1) = 0$

By the Zero Product Property:

$\displaystyle 2u + 1 = 0\text{ or } u + 1 = 0$

Solve for each case:

$\displaystyle u = -\frac{1}{2} \text{ or } u = -1$

Back-substitute:

$\displaystyle \sin \theta = -\frac{1}{2} \text{ or } \sin \theta =-1$

Use the Unit Circle. Hence, our three solutions in the interval [0, 2π) are:

$\displaystyle \theta = \left\{\frac{7\pi}{6}, \, \frac{3\pi}{2}, \frac{11\pi}{6}\right\}$