**Hint: **In this question, we are given the value of the cosine of the angle and the quadrant in which the angle x lies. Therefore, using the definition of cos(x), and other trigonometric formulas, we can obtain the values of other trigonometric ratios by solving the corresponding equations.

**Complete step by step solution:**

We are given the value of cot(x). We can use the trigonometric relation

${{\sin }^{2}}\left( x \right)=1-{{\cos }^{2}}\left( x \right)$

With the given value of cos(x) to find

$ {{\sin }^{2}}\left( x \right)=1-{{\left( \dfrac{-1}{2} \right)}^{2}}=1-\dfrac{1}{4}=\dfrac{4-1}{4}=\dfrac{3}{4} $

$ \Rightarrow \sin (x)=\pm \sqrt{\dfrac{3}{4}}=\pm \dfrac{\sqrt{3}}{2} $

However, in the third quadrant, the value of sin(x) is negative, so we should take only the negative value in the above equation to obtain

$\sin (x)=\dfrac{-\sqrt{3}}{2}............(1.1)$

Now, the expression for tan(x) is given by,

$\tan (x)=\dfrac{\sin (x)}{\cos (x)}$

Therefore, using the values of cos(x) and sin(x) from the values given in question and (1.1),

$\tan (x)=\dfrac{\sin (x)}{\cos(x)}=\dfrac{\dfrac{-1}{2}}{\dfrac{-\sqrt{3}}{2}}=\dfrac{1}{\sqrt{3}}............(1.2)$

Also tan(x) and cot(x) are related by

$\tan (x)=\dfrac{1}{\cot (x)}$

Therefore, using the value of tan(x) from (1.2) in the above equation, we find

$\cot (x)=\dfrac{1}{\dfrac{1}{\sqrt{3}}}=\sqrt{3}...............(1.3)$

We know that cosec(x) and sec(x) are the multiplicative inverses of sin(x) and cos(x) as,

therefore using the given value of cos(x) and by using equation (1.1), we obtain

$\text{cosec}(x)=\dfrac{1}{\sin(x)}=\dfrac{1}{\dfrac{-\sqrt{3}}{2}}=\dfrac{-2}{\sqrt{3}}..........(1.4)$

And

$\text{sec}(x)=\dfrac{1}{\cos (x)}=\dfrac{1}{\dfrac{-1}{2}}=-2...........(1.5)$

Therefore, from equations (1.1), (1.2), (1.3), (1.4) and (1.5), we have found the other trigonometric ratios as

$\sin (x)=\dfrac{-\sqrt{3}}{2}$, $\tan (x)=\dfrac{1}{\sqrt{3}}$, $\cot (x)=\sqrt{3}$ , $\sec(x)=-2$, $\text{cosec}(x)=\dfrac{-2}{\sqrt{3}}$

Which is the required answer to this question.

**Note: **We could also have found sec(x) by using $\sec (x)=\dfrac{1}{\cos (x)}$ and then tan(x) from sec(x) by using the identity ${{\sec }^{2}}\left( x \right)=1+{{\tan }^{2}}\left( x \right)$ and from there we could have found sin(x) and cos(x). However, the answer would have remained the same as found out in the solution above.

Find the values of other five trigonometric functions if $\cos (x)=\dfrac{-1}{2}$, x lies in the third quadrant.