Question

How do you find all solutions of the equation $4{\cos ^2}x - 3 = 0?$

Answer 1

Hint:First solve the given trigonometric equation for the value of $\cos x$, you will probably get two values for it then separately solve both the values of $\cos x$ for the value of $x$ and then to find all the solutions, find the general solution of $x$ for both values of cosine. And finally take the union of both values to get all solutions together.

Complete step by step answer:
In order to find all solutions of the given trigonometric equation $4{\cos ^2}x - 3 = 0$ we will first find the values we get for $\cos x$ from the given equation by solving it for $\cos x$ first.
$4{\cos ^2}x - 3 = 0 \\
\Rightarrow 4{\cos ^2}x = 3 \\
\Rightarrow {\cos ^2}x = \dfrac{3}{4} \\ $
Now if we take square root both sides, above equation will be
$\cos x = \pm \sqrt {\dfrac{3}{4}} \\
\Rightarrow \cos x = \pm \dfrac{{\sqrt 3 }}{2} \\ $
So we get two values of $\cos x$ that are $\cos x = \dfrac{{\sqrt 3 }}{2}\;{\text{and}}\; - \dfrac{{\sqrt 3 }}{2}$
Solving for $\cos x = \dfrac{{\sqrt 3 }}{2}$ first
$\cos x = \dfrac{{\sqrt 3 }}{2} \\
\Rightarrow x = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\ $
We know that the value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ is equal to $\dfrac{\pi }{6}$, and we also know that cosine function is positive in first and forth quadrant, so its general solution will be given as
$x = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\
\Rightarrow x = 2n\pi \pm \dfrac{\pi }{6} \\ $
Now solving for $\cos x = - \dfrac{{\sqrt 3 }}{2}$
$\cos x = - \dfrac{{\sqrt 3 }}{2} \\
\Rightarrow x = \cos \left( { - \dfrac{{\sqrt 3 }}{2}} \right) \\ $
We know that the value of ${\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$ is equal to $\dfrac{{7\pi }}{6}$, and we also know that cosine function is negative in second and third quadrant, so its general solution will be given as
$x = {\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) \\
\therefore x = 2n\pi \pm \dfrac{{7\pi }}{6} \\ $
Therefore the required solution for the given trigonometric equation will be given as $x = \left( {2n\pi \pm \dfrac{\pi }{6}} \right) \cup \left( {2n\pi \pm \dfrac{{7\pi }}{6}} \right)$.

Note:When solving the trigonometric equations like this in which we have to find the argument of the given trigonometric equation, always solve the equation first for the principle argument then find the respective general argument taking the quadrant and principle solution in note.