# How do you solve sin(2x-1)=sin(3x+1) ?

May 19, 2018

$x = - 2$

#### Explanation:

If $\sin \left(2 x - 1\right) = \sin \left(3 x + 1\right)$

Then it follows that

$\textcolor{w h i t e}{\text{.")2x-1color(white)("d")=color(white)("d}} 3 x + 1$

$- 1 - 1 \textcolor{w h i t e}{\text{d")=color(white)("d}} 3 x - 2 x$

$\textcolor{w h i t e}{\text{d.dd")-2color(white)("d")=color(white)("d}} x$

May 19, 2018

$x = \left\{\left(2 k + 1\right) \frac{\pi}{5} , k \in \mathbb{Z}\right\} \cup \left\{2 k \pi - 2 , k \in \mathbb{Z}\right\}$

#### Explanation:

Here,

$\sin \left(2 x - 1\right) = \sin \left(3 x + 1\right)$

$\implies \sin \left(3 x + 1\right) - \sin \left(2 x - 1\right) = 0$

We know that,

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

Using $\left(1\right)$ we get

$2 \cos \left(\frac{\left(3 x + 1\right) + \left(2 x - 1\right)}{2}\right) \sin \left(\frac{\left(3 x + 1\right) - \left(2 x - 1\right)}{2}\right) = 0$

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

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

$\left(i\right) \cos \left(\frac{5 x}{2}\right) = 0 \implies \frac{5 x}{2} = \left(2 k + 1\right) \frac{\pi}{2} , k \in \mathbb{Z}$

$\implies 5 x = \left(2 k + 1\right) \pi , k \in \mathbb{Z}$

=>color(blue)(x=(2k+1)pi/5, kin ZZ

$\left(i i\right) \sin \left(\frac{x + 2}{2}\right) = 0 \implies \frac{x + 2}{2} = k \pi , k \in \mathbb{Z}$

$\implies x + 2 = 2 k \pi , k \in \mathbb{Z}$

=>color(blue)(x=2kpi-2,kinZZ

Hence,

$x = \left\{\left(2 k + 1\right) \frac{\pi}{5} , k \in \mathbb{Z}\right\} \cup \left\{2 k \pi - 2 , k \in \mathbb{Z}\right\}$

May 19, 2018

$x = - {114}^{\circ} 65 + k {360}^{\circ}$
$x = {36}^{\circ} + k {360}^{\circ}$

#### Explanation:

sin (2x - 1) = sin (3x + 1)
Unit circle and property of sin function give 2 solutions for (2x - 1):
$\left(2 x - 1\right) = \left(3 x + 1\right)$ and $\left(2 x - 1\right) = \pi - \left(3 x + 1\right)$
a. (2x - 1) = (3x + 1)
$x = \frac{\left({180}^{\circ}\right) \left(- 2\right)}{3.14} = - {114}^{\circ} 65 + k {360}^{\circ}$
b. $\left(2 x - 1\right) = \pi - \left(3 x + 1\right) = \pi - 3 x - 1$
$5 x = \pi = {180}^{\circ}$
$x = \frac{180}{5} = {36}^{\circ} + k {360}^{\circ}$
Check with calculator.
x = - 114.65 --> 2x = - 229.30 --> 3x = - 343.95 -->
1 radians = 180/3.14 = 57^@32 -->
(2x - 1) = - 229.3 - 57.32 = - 286.62 --> sin (-286.62) = 0.96 -->
sin (3x + 1) = sin (- 343.95 + 57.32) = sin (-286.63) = 0.96. Proved.
x = 36 --> sin (2x - 1) = sin (72 - 57.32) = sin (14.68) = 0.25
sin (3x + 1) = sin (108 + 57.32) = sin (165.32) = 0.25. Proved