# How do you solve sin(2theta) + sin(4theta) = 0?

Apr 16, 2015

You can solve it with the formula sum-to-product, that says:

$\sin \alpha + \sin \beta = 2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$,

so:

$\sin 2 \theta + \sin 4 \theta = 0 \Rightarrow 2 \sin \left(\frac{2 \theta + 4 \theta}{2}\right) \cos \left(\frac{2 \theta - 4 \theta}{2}\right) = 0 \Rightarrow$

$\sin 3 \theta \cos \left(- \theta\right) = 0$

so:

$\sin 3 \theta = 0 \Rightarrow 3 \theta = k \pi \Rightarrow \theta = k \frac{\pi}{3}$

and:

$\cos \left(- \theta\right) = 0 \Rightarrow \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} + k \pi$.

(remembering that $\cos \left(- \alpha\right) = \cos \alpha$).

There is another way:

$\sin 2 \theta = - \sin 4 \theta \Rightarrow \sin 3 \theta = \sin \left(- 4 \theta\right)$

(remembering that $\sin \left(- \alpha\right) = - \sin \alpha$),

and than remembering that two sini are equal when the angles are equal or if the angles are supplementary,

$2 \theta = - 4 \theta + 2 k \pi \Rightarrow 6 \theta = 2 k \pi \Rightarrow \theta = k \frac{\pi}{3}$

and

$2 \theta = \pi - \left(- 4 \theta\right) + 2 k \pi \Rightarrow 2 \theta = \pi + 4 \theta + 2 k \pi \Rightarrow$

$- 2 \theta = \pi + 2 k \pi \Rightarrow \theta = - \frac{\pi}{2} - k \pi$ that is formally identical to the second solution found before.

Oct 22, 2016

$\theta = \frac{k \pi}{2} , \frac{\pi}{3} + k \pi , \frac{2 \pi}{3} + k \pi \text{ "," } k \in \mathbb{Z}$

#### Explanation:

The sine double-angle formula is $\sin \left(2 \alpha\right) = 2 \sin \left(\alpha\right) \cos \left(\alpha\right)$. Thus, $\sin \left(4 \theta\right) = 2 \sin \left(2 \theta\right) \cos \left(2 \theta\right)$. The given equation is then equivalent to

$\sin \left(2 \theta\right) + 2 \sin \left(2 \theta\right) \cos \left(2 \theta\right) = 0$

Factoring yields

$\sin \left(2 \theta\right) \left(1 + 2 \cos \left(2 \theta\right)\right) = 0$

Now, we can solve the two resulting equations whose product is $0$. The first gives

$\sin \left(2 \theta\right) = 0$

Thinking about when the sine function is $0$, we see that

$2 \theta = 0 , \pi , 2 \pi , 3 \pi \ldots$

Which can be generalized when $k$ is an integer by saying

$2 \theta = k \pi \text{ "," } k \in \mathbb{Z}$

Note that $k \in \mathbb{Z}$ is the symbolic way of representing that $k$ is an integer. Thus

color(blue)(theta=(kpi)/2" "," "kinZZ

The other resultant equation from before was

$1 + 2 \cos \left(2 \theta\right) = 0$

So

$\cos \left(2 \theta\right) = - \frac{1}{2}$

This happens at

$2 \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3}$

And all of these angles coterminal versions, which are located at integer multiples of $2 \pi$ away. Thus

$2 \theta = \frac{2 \pi}{3} + 2 k \pi , \frac{4 \pi}{3} + 2 k \pi \text{ "," } k \in \mathbb{Z}$

Solving for $\theta$ yields

color(blue)(theta=pi/3+kpi,(2pi)/3+kpi" "," "kinZZ

We can combine all of our solutions into

color(red)(theta=(kpi)/2,pi/3+kpi,(2pi)/3+kpi" "," "kinZZ