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**Hint:**We will put the given value of cosine in the formula and then we will solve the question. On doing some simplification we get the required answer.

**Formula used:**\[\cos 2x\] have four different kinds of formulas.

\[(1)\;\cos 2x = {\cos ^2}x - {\sin ^2}x\]

Now, we know that \[{\cos ^2}x + {\sin ^2}x = 1\].

Using this formula we can derive the following formula also:

\[(2)\;\;\operatorname{Cos} 2x = 1 - 2{\operatorname{Sin} ^2}x\]

Now, using the \[{\cos ^2}x + {\sin ^2}x = 1\], we can convert the \[\cos 2x\]in following way:

\[(3)\;\;\cos 2x = 2{\cos ^2}x - 1\]

\[(4)\;\;\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}\].

**Complete Step by Step Solution:**

In the above question, the value of ‘\[\cos x\]’ is given.

So, we will put the value of ‘\[\cos x\]’ directly in the formula to derive the value of ‘\[\cos 2x\]’.

So, we will put the value of \[\cos x = \dfrac{5}{{13}}\] in the third formula.

So, we will put \[\cos x = \dfrac{5}{{13}}\]in \[\cos 2x = 2{\cos ^2}x - 1\].

So, after putting this value, we can rewrite the equation in following way:

\[ \Rightarrow \cos 2x = 2 \times {\left( {\dfrac{5}{{13}}} \right)^2} - 1\].

Now, squaring the constant term, we get the following value:

\[ \Rightarrow \cos 2x = 2 \times \left( {\dfrac{{25}}{{169}}} \right) - 1\].

Now, multiply the first two terms, we get the following expression:

\[ \Rightarrow \cos 2x = \dfrac{{50}}{{169}} - 1\].

Now, subtract the above terms, we get the following value:

\[ \Rightarrow \cos 2x = \dfrac{{50 - 169}}{{169}}\].

Now, subtract the terms in numerator, we get the following value:

\[ \Rightarrow \cos 2x = \dfrac{{ - 119}}{{\;\;169}}\].

Now, we can rewrite the above expression in following way:

\[ \Rightarrow \cos 2x = - \dfrac{{119}}{{169}}\].

**Therefore, the exact value of \[\cos 2x\] is \[ - \dfrac{{119}}{{169}}\].**

**Note:**Points to remember:

A reflex angle is one that lies between \[{180^ \circ }\] and \[{360^ \circ }\].

It means that it will be either in the \[III\]quadrant or in \[IV\] quadrant.

Recall that for \[{\text{cos}}\theta = \dfrac{{{\text{adjacent}}}}{{{\text{hypotenuse}}}}\].

Hypotenuse is never negative, so the adjacent must be the negative one.

If the adjacent is the negative one, it must be in the \[III\] quadrant.

This also means that the opposite must be negative as well.

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