How do you solve $ 3{\sec ^2}x - 4 = 0 $ ?

Hint: In order to solve this question ,transpose everything from left-Hand side to right-hand side except $ {\sec ^2}x $ and then put $ {\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}} $ determine the angle whose cosine is equivalent to \[\dfrac{{\sqrt 3 }}{2}\] to get the set of desired solutions.
Formula:
 $ \sin \left( {A - B} \right) = \sin \left( A \right)\cos \left( B \right) - \sin \left( B \right)\cos \left( A \right) $

Complete step-by-step answer:
We are given a trigonometric expression $ 3{\sec ^2}x - 4 = 0 $
 $
   \Rightarrow 3{\sec ^2}x - 4 = 0 \\
   \Rightarrow 3{\sec ^2}x = 4 \\
   \Rightarrow {\sec ^2}x = \dfrac{4}{3} \;
  $
As we know that in trigonometry $ {\sec ^2}x = \dfrac{1}{{{{\cos }^2}x}} $ ,putting the above expression
 $ \Rightarrow \dfrac{1}{{{{\cos }^2}x}} = \dfrac{4}{3} $
Taking reciprocal on both the sides.
 \[
   \Rightarrow {\cos ^2}x = \dfrac{3}{4} \\
   \Rightarrow \cos x = \sqrt {\dfrac{3}{4}} \\
   \Rightarrow \cos x = \dfrac{{\sqrt 3 }}{{\sqrt 4 }} \\
   \Rightarrow \cos x = \dfrac{{\sqrt 3 }}{2}' \\
   \Rightarrow x = {\cos ^{ - 1}}\dfrac{{\sqrt 3 }}{2} \;
 \]
 \[{\cos ^{ - 1}}\dfrac{{\sqrt 3 }}{2}\] = An angle whose cosine is equal to \[\dfrac{{\sqrt 3 }}{2}\] .
Hence, \[x = \dfrac{\pi }{6} + 2n\pi \] where $ n \in Z $ , $ Z $ represents the set of all integers.
Therefore, the solution to expression $ 3{\sec ^2}x - 4 = 0 $ is \[x = \dfrac{\pi }{6} + 2n\pi \] where $ n \in Z $ , $ Z $ represents the set of all integers.
So, the correct answer is “ \[x = \dfrac{\pi }{6} + 2n\pi \] ”.

Note: 1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
3. $ Z $ represents the set of all integers.