# How to solve sin5x = -sin3x?

Apr 23, 2018

$x = \left\{\left(2 k + 1\right) \frac{\pi}{2} , k \in \mathbb{Z}\right\} U \left\{\frac{k \pi}{4} , k \in \mathbb{Z}\right\}$

#### Explanation:

We know that,

color(red)((1)sinC+sinD=2sin((C+D)/2)cos((C-D)/2)

color(blue)((2)sintheta=0=>theta=kpi,kinZZ

color(blue)((3)costheta=0=>theta=(2k+1)pi/2,kinZZ

Here,

$\sin 5 x = - \sin 3 x$

=>sin5x+sin3x=0...tocolor(red)(Apply(1)

$\implies 2 \sin \left(\frac{5 x + 3 x}{2}\right) \cos \left(\frac{5 x - 3 x}{2}\right) = 0$

$\implies \sin 4 x \cos x = 0$

$\implies \sin 4 x = 0 \mathmr{and} \cos x = 0$

(i)sin4x=0=>4x=kpi,kinZZ...tocolor(blue)(Apply(2)

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots} \implies x = \frac{k \pi}{4} , k \in \mathbb{Z}$

(ii)cosx=0=>x=(2k+1)pi/2,kinZZ...tocolor(blue)(Apply(3)

Hence, from $\left(i\right) \mathmr{and} \left(i i\right)$

$x = \left\{\frac{k \pi}{4} , k \in \mathbb{Z}\right\} U \left\{\left(2 k + 1\right) \frac{\pi}{2} , k \in \mathbb{Z}\right\}$

Apr 23, 2018

$x \in \left\{\frac{1}{4} k \pi : k \in \mathbb{Z}\right\} \cup \left\{\frac{2 k + 1}{2} \pi : k \in \mathbb{Z}\right\}$

#### Explanation:

We have the following properties of $\sin$ which we will use to answer the question:

• $- \sin x = \sin \left(- x\right)$
• $\sin x = \sin \left(\pi - x\right)$

Let $k$ be an integer.

$\sin 5 x = - \sin 3 x = \sin \left(- 3 x\right) = \sin \left(\pi + 3 x\right)$

$5 x = - 3 x + 2 k \pi$ or $5 x = \pi + 3 x + 2 k \pi$

so $8 x = 2 k \pi$ or $2 x = \pi + 2 k \pi$

so $x = \frac{1}{4} k \pi$ or $x = \frac{1}{2} \pi + k \pi = \frac{2 k + 1}{2} \pi$

so $x \in \left\{\frac{1}{4} k \pi : k \in \mathbb{Z}\right\} \cup \left\{\frac{2 k + 1}{2} \pi : k \in \mathbb{Z}\right\}$