# Solve the equation 2sin3x+sqrt3=0?

Mar 23, 2017

General solution is $x = \frac{n \pi}{3} - {\left(- 1\right)}^{n} \frac{\pi}{9}$

#### Explanation:

As $2 \sin 3 x + \sqrt{3} = 0$

$\sin 3 x = - \frac{\sqrt{3}}{2} = - \sin \left(\frac{\pi}{3}\right)$

Hence General Solution is $3 x = n \pi + {\left(- 1\right)}^{n} \left(- \frac{\pi}{3}\right)$

or $3 x = n \pi - {\left(- 1\right)}^{n} \frac{\pi}{3}$

and $x = \frac{n \pi}{3} - {\left(- 1\right)}^{n} \frac{\pi}{9}$

and a few solutions using $n = 0 , 1 , 2 , 3 , 4 , \ldots \ldots . .$ are

$x = - \frac{\pi}{9} , \frac{4 \pi}{9} , \frac{5 \pi}{9} , \frac{4 \pi}{3} , \ldots \ldots \ldots \ldots . .$

Mar 24, 2017

see below

#### Explanation:

$2 \sin 3 x + \sqrt{3} = 0$

$2 \sin 3 x = - \sqrt{3}$

$\sin 3 x = - \frac{\sqrt{3}}{2}$

$3 x = {\sin}^{-} 1 \left(- \frac{\sqrt{3}}{2}\right)$

$3 x = - \frac{\pi}{3} + 2 \pi n , \mathmr{and} 3 x = - \frac{2 \pi}{3} + 2 \pi n$

$x = \left(- \frac{\pi}{3} + 2 \pi n\right) \left(\frac{1}{3}\right) , \mathmr{and} x = \left(- \frac{2 \pi}{3} + 2 \pi n\right) \left(\frac{1}{3}\right)$

$x = - \frac{\pi}{9} + \frac{2 \pi}{3} n , \mathmr{and} x = - \frac{2 \pi}{9} + \frac{2 \pi}{3} n$

$S = \left\{- \frac{\pi}{9} + \frac{2 \pi}{3} n , - \frac{2 \pi}{9} + \frac{2 \pi}{3} n\right\}$