How do you verify that the $ x $ values $ \dfrac{\pi }{{16}} $ and $ \dfrac{{3\pi }}{{16}} $ are solutions to $ 2{\cos ^2}4x - 1 = 0 $ ?

Hint: Here, you are given two values of $ x $ and a trigonometric equation and you are asked to verify the solutions. So, whenever you are given a function $ f\left( x \right) = 0 $ , you solve the equation and find the value of $ x $ and that will be your solution. As you know that cosine is a periodic function, you need to think of a general solution for the above given equation.

Complete step by step solution:
The given to you is $ 2{\cos ^2}4x - 1 = 0 $ . No, what we will do is, try to solve and obtain values of $ x $ that will satisfy the equation. Let us add $ 1 $ on both side of the equation, we will get,
$
2{\cos ^2}4x - 1 + 1 = 0 + 1 \\
2{\cos ^2}4x = 1 \;
 $
Now, dividing both sides of the equation by $ 2 $ , we will get,
 $
  \dfrac{{2{{\cos }^2}4x}}{2} = \dfrac{1}{2} \\
  {\cos ^2}4x = \dfrac{1}{2} \;
 $
Now, take square on both sides of the equation, we will get,
\[
  \sqrt {{{\cos }^2}4x} = \sqrt {\dfrac{1}{2}} \\
   \pm \cos 4x = \dfrac{1}{{\sqrt 2 }} \\
   \to \cos 4x = \pm \dfrac{1}{{\sqrt 2 }} \;
 \]
Here, after taking the square root of $ {\cos ^2}4x $ , you get two values of $ \cos 4x $ , one will be positive and the other will be negative.
The solution for $ \cos x = \cos \theta $ is $ x = 2n\pi \pm \theta $ . In our case, $ \theta = \dfrac{\pi }{4} $ and \[\theta = \dfrac{{3\pi }}{4}\] because $ \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} $ and $ \cos \dfrac{{3\pi }}{4} = - \dfrac{1}{{\sqrt 2 }} $ , also $ x \to 4x $ . Therefore, $ 4x = 2n\pi \pm \dfrac{\pi }{4} $ and $ 4x = 2n\pi \pm \dfrac{{3\pi }}{4} $
\[
  4x = 2n\pi \pm \dfrac{\pi }{4} \\
\Rightarrow x = \dfrac{{2n\pi \pm \dfrac{\pi }{4}}}{4} \\
\Rightarrow x = \dfrac{{n\pi }}{2} \pm \dfrac{\pi }{{16}} \\
 \]
and
 \[
  4x = 2n\pi \pm \dfrac{{3\pi }}{4} \\
\Rightarrow x = \dfrac{{2n\pi \pm \dfrac{{3\pi }}{4}}}{4} \\
\Rightarrow x = \dfrac{{n\pi }}{2} \pm \dfrac{{3\pi }}{{16}} \;
 \]
These are the general solutions of our \[x\]. Here, $ n $ is an integer, that is, $ n = 0, \pm 1, \pm 2, \pm 3,... $ as integer numbers include both positive numbers and negative numbers. Now, if you consider the solution which we obtained by solving the given equation, that is
\[x = \dfrac{{n\pi }}{2} \pm \dfrac{\pi }{{16}}\] and \[x = \dfrac{{n\pi }}{2} \pm \dfrac{{3\pi }}{{16}}\] ,
if you put $ n = 0 $ , you will get $ x = \dfrac{\pi }{{16}},\dfrac{{3\pi }}{{16}} $ .
So, yes, the $ x $ values $ \dfrac{\pi }{{16}} $ and $ \dfrac{{3\pi }}{{16}} $ are solutions to
$ 2{\cos ^2}4x - 1 = 0 $ .
Hence, $ x $ values $ \dfrac{\pi }{{16}} $ and $ \dfrac{{3\pi }}{{16}} $ are solutions to $ 2{\cos ^2}4x - 1 = 0 $ , verified.

Note: Here, we have considered the general solution of $ \cos x = \cos \theta $ which is $ x = 2n\pi \pm \theta $ , you need to memorize this. Also, you need to remember the general solutions of three trigonometric ratios, that are sine, cosine and tangent. Whenever you are asked to verify the solution, just equate the general solution with the given solution and find the value of $ n $ , if it comes out to be an integer, then the given value if the solution of the provided equation, else it is not a solution.