Question

Can someone show me how to factor this please?

Answer 1

`\displaystyle \theta = \left\{(7\pi)/(6), \, (3\pi)/(2), (11\pi)/(6)\right\}`

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 = -(1)/(2) \text{ or } u = -1`

Back-substitute:

`\displaystyle \sin \theta = -(1)/(2) \text{ or } \sin \theta =-1`

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

`\displaystyle \theta = \left\{(7\pi)/(6), \, (3\pi)/(2), (11\pi)/(6)\right\}`