# How do I solve this problem?

## Solve the equation by first using a sum to product formula: $\sin 4 \theta - \sin 2 \theta = \cos 3 \theta$

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#### Explanation

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#### Explanation:

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Noah G Share
Feb 9, 2018

I got $x = \frac{\pi}{6} + \frac{\pi}{2} n$ and $x = \frac{\pi}{2} + \pi n$

#### Explanation:

Let's start by rewriting:

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

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

$2 \sin \left(2 \theta\right) \cos \left(2 \theta\right) - \sin \left(2 \theta\right) = \cos \left(2 \theta\right) \cos \theta - \sin \left(2 \theta\right) \sin \theta$

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

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

Now recall that $\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \theta$.

$\frac{\left(2 \sin \theta \cos \theta\right) \left(2 \cos 2 \theta - 1\right)}{\cos} \theta = 1 - 4 {\sin}^{2} \theta$

$2 \sin \theta \left(2 \left(1 - 2 {\sin}^{2} \theta\right) - 1\right) = 1 - 4 {\sin}^{2} \theta$

$2 \sin \theta \left(2 - 4 {\sin}^{2} \theta - 1\right) = 1 - 4 {\sin}^{2} \theta$

$4 \sin \theta - 8 {\sin}^{3} \theta - 2 \sin \theta = 1 - 4 {\sin}^{2} \theta$

$0 = 8 {\sin}^{3} \theta - 4 {\sin}^{2} \theta - 2 \sin \theta + 1$

Now let $y = \sin \theta$.

$0 = 8 {y}^{3} - 4 {y}^{2} - 2 y + 1$

$0 = 4 {y}^{2} \left(2 y - 1\right) - \left(2 y - 1\right)$

$0 = \left(4 {y}^{2} - 1\right) \left(2 y - 1\right)$

${y}^{2} = \frac{1}{4} \mathmr{and} y = \frac{1}{2}$

$y = \pm \frac{1}{2}$

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

$\theta = \frac{\pi}{6} + \frac{\pi}{2} n$

However, we must account for when we cancelled the cosine above. This can be a dangerous game, because you can lose solutions. However, if you recognize it, you can prevent that from happening.

Since the equation at this point is a product, the equation will hold true whenever $\cos \theta = 0$ (rendering the entire equation $0$).

Thus, $\theta = \frac{\pi}{2} + \pi n$. The graph of the functions ${y}_{1} = \sin \left(4 x\right) - \sin \left(2 x\right)$ and ${y}_{2} = \cos \left(3 x\right)$ confirm that the solutions we found are indeed correct.

Hopefully this helps!

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