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Elenna [48]
3 years ago
9

Can someone show me how to factor this please?

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
8 0

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\}

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