How do you solve the following equation ${(\cot (t))^2} = \dfrac{3}{4}$ in the interval $\left[ {0,2\pi } \right]$?

Hint: In this question, we want to find the trigonometry angle value of the given equation between the intervals 0 to $2\pi $. First, we apply the square-root of both sides. Then determine the expression with the help of the basic acute angle. Based on that, we will be able to find the value of the angle. Find the angles in all four quadrants because the given interval is $\left[ {0,2\pi } \right]$.

Complete step-by-step answer:
In this question, given that
$ \Rightarrow {(\cot (t))^2} = \dfrac{3}{4}$
To simplify the above expression, let us apply the square root on both sides.
$ \Rightarrow \cot \left( t \right) = \pm \sqrt {\dfrac{3}{4}} $
That is equal to
$ \Rightarrow \cot \left( t \right) = \pm \dfrac{{\sqrt 3 }}{2}$
Let us determine the basic acute angle.
$ \Rightarrow t = \operatorname{arccot} \left( {\dfrac{{\sqrt 3 }}{2}} \right)$
The cosine function value is $\dfrac{{\sqrt 3 }}{2}$ at the angle of 30 in the first quadrant. In trigonometry angle 30 is also written as$\dfrac{\pi }{6}$. Same way, we can find all the values.
$ \Rightarrow t = \dfrac{\pi }{6}$
The interval is given as $\left[ {0,2\pi } \right]$. So the value for t will lie in all four quadrants.
This common value that we get with,
$ \Rightarrow t = \dfrac{\pi }{6},\left( {\pi - \dfrac{\pi }{6}} \right),\left( {\pi + \dfrac{\pi }{6}} \right),\left( {2\pi - \dfrac{\pi }{6}} \right)$
Let us take the least common factor in the above expression.
$ \Rightarrow t = \dfrac{\pi }{6},\left( {\dfrac{{6\pi - \pi }}{6}} \right),\left( {\dfrac{{6\pi + \pi }}{6}} \right),\left( {\dfrac{{12\pi - \pi }}{6}} \right)$
Simplify the above expression by applying addition and subtraction to the numerator.
So, the answer is
$ \Rightarrow t = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}$
We can also write the answer in set form. So, the solution set
$ \Rightarrow S = \left\{ {\dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}} \right\}$

Note:
Here, we must remember the trigonometry ratios and the value of ratio at the angles 0, 30, 45, 60, and 90. We also have to learn about the values in all four quadrants with a positive and negative sign.
The value of $\sin 0^\circ = 0$
The value of $\sin 30^\circ = \dfrac{1}{2}$
The value of $\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}$
The value of $\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$
The value of $\sin 90^\circ = 1$
The value of $\cos 0^\circ = 1$
The value of $\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}$
The value of $\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$
The value of $\cos 60^\circ = \dfrac{1}{2}$
The value of $\cos 90^\circ = 0$
The value of $\tan 0^\circ = 0$
The value of $\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$
The value of $\tan 45^\circ = 1$
The value of $\tan 60^\circ = \sqrt 3 $
The value of $\tan 90^\circ $ is not defined.