Answer
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Hint: We first simplify the equation $3\cot 2x-\sqrt{3}=0$ to find the value of $\cot 2x$. Then we find the principal value of x for which $3\cot 2x-\sqrt{3}=0$. In that domain, equal value of the same ratio gives equal angles. We find the angle value for x. At the end we also find the general solution for the equation $3\cot 2x-\sqrt{3}=0$.
Complete step-by-step solution:
It’s given that $3\cot 2x-\sqrt{3}=0$. We simplify the equation to get
$\begin{align}
& 3\cot 2x-\sqrt{3}=0 \\
& \Rightarrow \cot 2x=\dfrac{\sqrt{3}}{3}=\dfrac{1}{\sqrt{3}} \\
\end{align}$
The value in fraction is $\dfrac{1}{\sqrt{3}}$. We need to find x for which $\cot 2x=\dfrac{1}{\sqrt{3}}$.
We know that in the principal domain or the periodic value of $0\le x\le \pi $ for $\sin x$, if we get $\cot a=\cot b$ where $0\le a,b\le \pi $ then $a=b$.
We have the value of $\cot \left( \dfrac{\pi }{3} \right)$ as $\dfrac{1}{\sqrt{3}}$. $0<\dfrac{\pi }{3}<\pi $.
Therefore, $\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}=\cot \left( \dfrac{\pi }{3} \right)$ which gives $2x=\dfrac{\pi }{3}$.
For $\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}$, the value of x is $x=\dfrac{\pi }{6}$.
We also can show the solutions (primary and general) of the equation $\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}$ through the graph. We take $y=\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}$. We got two equations $y=\cot \left( 2x \right)$ and $y=\dfrac{1}{\sqrt{3}}$. We place them on the graph and find the solutions as their intersecting points.
We can see the primary solution in the interval $0\le x\le \pi $ is the point A as $x=\dfrac{\pi }{6}$.
All the other intersecting points of the curve and the line are general solutions.
Note: Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $0\le x\le \pi $. In that case we have to use the formula $x=n\pi +a$ for $\cot \left( x \right)=\cot a$ where $0\le a\le \pi $. For our given problem $\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}$, the general solution will be $2x=n\pi +\dfrac{\pi }{3}$. Here $n\in \mathbb{Z}$.
The simplified form of the general solution will be \[x=\dfrac{n\pi }{2}+\dfrac{\pi }{6}\].
Complete step-by-step solution:
It’s given that $3\cot 2x-\sqrt{3}=0$. We simplify the equation to get
$\begin{align}
& 3\cot 2x-\sqrt{3}=0 \\
& \Rightarrow \cot 2x=\dfrac{\sqrt{3}}{3}=\dfrac{1}{\sqrt{3}} \\
\end{align}$
The value in fraction is $\dfrac{1}{\sqrt{3}}$. We need to find x for which $\cot 2x=\dfrac{1}{\sqrt{3}}$.
We know that in the principal domain or the periodic value of $0\le x\le \pi $ for $\sin x$, if we get $\cot a=\cot b$ where $0\le a,b\le \pi $ then $a=b$.
We have the value of $\cot \left( \dfrac{\pi }{3} \right)$ as $\dfrac{1}{\sqrt{3}}$. $0<\dfrac{\pi }{3}<\pi $.
Therefore, $\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}=\cot \left( \dfrac{\pi }{3} \right)$ which gives $2x=\dfrac{\pi }{3}$.
For $\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}$, the value of x is $x=\dfrac{\pi }{6}$.
We also can show the solutions (primary and general) of the equation $\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}$ through the graph. We take $y=\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}$. We got two equations $y=\cot \left( 2x \right)$ and $y=\dfrac{1}{\sqrt{3}}$. We place them on the graph and find the solutions as their intersecting points.
We can see the primary solution in the interval $0\le x\le \pi $ is the point A as $x=\dfrac{\pi }{6}$.
All the other intersecting points of the curve and the line are general solutions.
Note: Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $0\le x\le \pi $. In that case we have to use the formula $x=n\pi +a$ for $\cot \left( x \right)=\cot a$ where $0\le a\le \pi $. For our given problem $\cot \left( 2x \right)=\dfrac{1}{\sqrt{3}}$, the general solution will be $2x=n\pi +\dfrac{\pi }{3}$. Here $n\in \mathbb{Z}$.
The simplified form of the general solution will be \[x=\dfrac{n\pi }{2}+\dfrac{\pi }{6}\].