# How do you solve sin(2x) - sin(3x) + sin(4x) = 0?

##### 1 Answer
May 11, 2015

The answer is $\left\{x \in \mathbb{R} : x = k \pi \mathmr{and} x = \frac{2 \pi}{3} + 2 k \pi \mathmr{and} x = - \frac{2 \pi}{3} + 2 k \pi , k \in \mathbb{Z}\right\}$

Remember:
$\sin \left(4 x\right) = 2 \sin \left(2 x\right) \cos \left(2 x\right)$
$\sin \left(3 x\right) = \sin \left(x\right) \cos \left(2 x\right) + \cos \left(x\right) \sin \left(2 x\right)$

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

Remember: $1 = 2 - 1$

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

Remember: $1 + \cos \left(2 x\right) = 2 {\cos}^{2} \left(x\right)$

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

We have one set of solutions: $A = \left\{x \in \mathbb{R} : \sin \left(x\right) = 0\right\} = {\left\{k \pi\right\}}_{k \in \mathbb{Z}}$

We now can simplify

$2 \cos \left(x\right) \left(4 {\cos}^{2} \left(x\right) - 1 - \cos \left(x\right)\right) = \cos \left(2 x\right)$
Remember: $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right) = 2 {\cos}^{2} \left(x\right) - 1$
$8 {\cos}^{3} \left(x\right) - 2 \cos \left(x\right) - 2 {\cos}^{2} \left(x\right) = 2 {\cos}^{2} \left(x\right) - 1$

$y = \cos \left(x\right)$

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

We notice it's

${\left(2 y\right)}^{3} - {\left(2 y\right)}^{2} - \left(2 y\right) + 1 = 0$
So $2 y = - 1$ and it's the sole solution in $\mathbb{R}$ (we can divide with Ruffini's rule, or calculate the local minimum to prove it)

So we have another set of solutions $B = \left\{x \in \mathbb{R} : \cos \left(x\right) = - \frac{1}{2}\right\} = \left\{\frac{2 \pi}{3} + 2 k \pi\right\} \cup \left\{- \frac{2 \pi}{3} + 2 k \pi\right\}$

We know we don't have any other solution so the set is $A \cup B$