# Cos 2aplha+7sinalpha=-3?

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Aviv S. Share
Feb 9, 2018

$\alpha = \frac{7 \pi}{6} + 2 \pi k , \frac{11 \pi}{6} + 2 \pi k$

#### Explanation:

$\cos \left(2 \alpha\right) + 7 \sin \left(\alpha\right) = - 3$

Use the identity $\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \theta$:

$1 - 2 {\sin}^{2} \alpha + 7 \sin \alpha = - 3$

$- 2 {\left(\sin \alpha\right)}^{2} + 7 \sin \alpha + 4 = 0$

Now, let $u = \sin \alpha$.

$- 2 {u}^{2} + 7 u + 4 = 0$

${u}^{2} - \frac{7}{2} u - 2 = 0$

$\left(u - 4\right) \left(u + \frac{1}{2}\right) = 0$

$u = - \frac{1}{2} , 4$

Now substitute $\sin \alpha$ back in for $u$.

$\sin \alpha = - \frac{1}{2} , \textcolor{red}{\cancel{\textcolor{b l a c k}{\sin \alpha = 4}}}$

The reason why $\sin \alpha = 4$ is crossed out is that there is no solution for it.

$\sin \alpha = - \frac{1}{2}$

This is true when $\alpha$ is at $\frac{7 \pi}{6}$ and $\frac{11 \pi}{6}$ (and if you add any multiple of $2 \pi$ to either of them). The final answer set is:

$\alpha = \frac{7 \pi}{6} + 2 \pi k , \frac{11 \pi}{6} + 2 \pi k$

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