Question

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

Answer 1

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.