Answer
The exact solution of the equation $2\sin 3x+\sqrt{3}=0$ on the interval $\left[ 0,2\pi \right)$ is $\frac{4\pi }{9}$ , $\frac{5\pi }{9}$ , $\frac{10\pi }{9}$ , $\frac{11\pi }{9}$ , $\frac{16\pi }{9}$ , and $\frac{17\pi }{9}$.
Work Step by Step
Let us consider the provided expression:
$2\sin 3x+\sqrt{3}=0$
Add $-\sqrt{3}\text{ }$ on both sides:
$\begin{align}
& 2\sin 3x+\sqrt{3}-\sqrt{3}=0-\sqrt{3} \\
& 2\sin 3x=-\sqrt{3}
\end{align}$
And divide by $2$ on both sides:
$\begin{align}
& \frac{2\sin 3x}{2}=\frac{-\sqrt{3}}{2} \\
& \sin 3x=\frac{-\sqrt{3}}{2}
\end{align}$
On, the interval $\left[ 0,2\pi \right),$ the sine function is $\frac{-\sqrt{3}}{2}$ at $\frac{4\pi }{3}\text{ and }\frac{5\pi }{3}$.
So,
$\begin{align}
& \sin 3x=\frac{-\sqrt{3}}{2} \\
& 3x=\frac{4\pi }{3}
\end{align}$
Or,
$\begin{align}
& \sin 3x=\frac{-\sqrt{3}}{2} \\
& 3x=\frac{5\pi }{3}
\end{align}$
Since the period of the sine function is $2\pi $, the general solution of the equation is:
$\begin{align}
& 3x=\frac{4\pi }{3}+2n\pi \\
& x=\frac{4\pi }{9}+\frac{2n\pi }{3}
\end{align}$
Or,
$\begin{align}
& 3x=\frac{5\pi }{3}+2n\pi \\
& x=\frac{5\pi }{9}+\frac{2n\pi }{3}
\end{align}$
To get different solutions, put $\text{ }n=0,1,2,3\ldots $
When $n=0$ ,
$x=\frac{4\pi }{9}\text{ or }x=\frac{5\pi }{9}$
When $n=1$ ,
$\begin{align}
& x=\frac{4\pi }{9}+\frac{2\pi }{3} \\
& =\frac{10\pi }{9}
\end{align}$
Or,
$\begin{align}
& x=\frac{5\pi }{9}+\frac{2\pi }{3} \\
& =\frac{11\pi }{9}
\end{align}$
When $n=2$ ,
$\begin{align}
& x=\frac{4\pi }{9}+\frac{4\pi }{3} \\
& =\frac{16\pi }{9}
\end{align}$
Or,
$\begin{align}
& x=\frac{5\pi }{9}+\frac{4\pi }{3} \\
& =\frac{17\pi }{9}
\end{align}$
Put the other values of $n$; then the solution becomes outside the interval $\left[ 0,2\pi \right).$
Thus, the exact solution of the equation $2\sin 3x+\sqrt{3}=0$ on the interval $\left[ 0,2\pi \right)$ is $\frac{4\pi }{9}$ , $\frac{5\pi }{9}$ , $\frac{10\pi }{9}$ , $\frac{11\pi }{9}$ , $\frac{16\pi }{9}$ , and $\frac{17\pi }{9}$.
