How do you use double angle formulas to calculate $\operatorname{Cos} 2x$ and $\operatorname{Sin} 2x$ without finding $x$ if $\operatorname{Cos} x = \dfrac{3}{5}$ and $x$ is in the first quadrant?

Hint: Formulas which represent trigonometric functions of an angle $2\theta $in the form of a trigonometric function of an angle $\theta $ are known as the double angle formulas.
The basic double angle formulas are:
$
  sin(2\theta ) = 2\sin \theta .\cos \theta \\
  cos(2\theta ) = {\cos ^2}\theta - {\sin ^2}\theta \\
  \tan (2\theta ) = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} \\
 $

Complete step by step answer:
$\cos x = \dfrac{3}{5}$
Now, we know that ${\sin ^2}x + {\cos ^2}x = 1$
Hence,
$
  {\sin ^2}x = 1 - {\cos ^2}x \\
  \sin x = \sqrt {1 - {{\cos }^2}x} \\
 $
It’s mentioned that $x$ is in first quadrant, Hence, $\sin x$and$\cos x$will be a positive value.
Now, substituting the values we will get,
$
  \sin x = \sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}} \\
  \sin x = \sqrt {1 - \dfrac{9}{{25}}} \\
  \sin x = \sqrt {\dfrac{{25 - 9}}{{25}}} \\
  \sin x = \sqrt {\dfrac{{16}}{{25}}} = \dfrac{4}{5} \\
 $
 Substituting these values to the double angle formulas, we get,
$
  \sin 2x = 2\sin x.\cos x \\
  \sin 2x = 2.\dfrac{4}{5}.\dfrac{3}{5} \\
  \sin 2x = \dfrac{{24}}{{25}} \\
 $
And
$
  \cos 2x = {\cos ^2}x - {\sin ^2}x \\
  \cos 2x = \dfrac{9}{{25}} - \dfrac{{16}}{{25}} \\
  \cos 2x = \dfrac{{ - 7}}{{25}} \\
 $
Hence, in this way we are able to calculate the value of $\sin 2x$ and $\cos 2x$ without finding the value of $x$.

Note: The sum and differences of sine and cosine formulas produce double angle and half angle formulas. That is the formula is derived by expanding $\sin (\theta + \theta )$ and by expanding $\cos (\theta + \theta )$.