# How do you solve tantheta = sqrt(2)sintheta?

Mar 31, 2017

$\theta = \left(\pm k \pi \cup \pm \frac{\pi}{4} + 2 k \pi\right)$ for $k \in \mathbb{Z}$

#### Explanation:

$\tan \left(\theta\right) = \sqrt{2} \sin \left(\theta\right)$ or

$\tan \left(\theta\right) \left(1 - \sqrt{2} \cos \left(\theta\right)\right) = 0$

$\left\{\begin{matrix}\tan \left(\theta\right) = 0 \to \theta = \pm k \pi \\ \cos \left(\theta\right) = \frac{1}{\sqrt{2}} \to \theta = \pm \frac{\pi}{4} + 2 k \pi\end{matrix}\right.$

$\theta = \left(\pm k \pi \cup \pm \frac{\pi}{4} + 2 k \pi\right)$ for $k \in \mathbb{Z}$

Mar 31, 2017

$\theta = \left\{\frac{\pi}{4} , \frac{7 \pi}{4}\right\}$

#### Explanation:

Here's an alternative approach. Note that $\tan \theta = \sin \frac{\theta}{\cos} \theta$.

$\sin \frac{\theta}{\cos} \theta = \sqrt{2} \sin \theta$

Multiply both sides by $\cos \theta$.

$\sin \frac{\theta}{\cos} \theta \cdot \cos \theta = \sqrt{2} \sin \theta \cdot \cos \theta$

Now recognize that $\sin 2 \theta = 2 \sin \theta \cos \theta$.

$\sin \theta = \sqrt{2} \sin \theta \cos \theta$

Multiply the right side by $\frac{2}{2}$
$\sin \theta = 2 \sqrt{\frac{1}{2}} \sin \theta \cos \theta$

$\sin \theta = \sqrt{\frac{1}{2}} \sin 2 \theta$

$\sin \frac{\theta}{2 \sin \theta \cos \theta} = \frac{1}{\sqrt{2}}$

$\frac{1}{2 \cos \theta} = \frac{1}{\sqrt{2}}$

$\sqrt{2} = 2 \cos \theta$

$\frac{\sqrt{2}}{2} = \cos \theta$

This is the rationalized form of $\frac{1}{\sqrt{2}}$.

$\frac{1}{\sqrt{2}} = \cos \theta$

This has solutions $\theta = \frac{\pi}{4} , \frac{7 \pi}{4}$.

Hopefully this helps!