# How do you solve cos^2x=3/4? Which solution is correct?

## ${\cos}^{2} x = \frac{3}{4}$ This is already covered here: https://socratic.org/questions/how-do-you-solve-cos-2x-3-4 But my solution: 1. $\cos x = \frac{\sqrt{3}}{2}$ 2. $\cos x = - \frac{\sqrt{3}}{2}$ So: 1. $\arccos \left(\frac{\sqrt{3}}{2}\right) + 2 \pi k = \frac{\pi}{6} + 2 \pi k$ $- \arccos \left(\frac{\sqrt{3}}{2}\right) + 2 \pi k = - \frac{\pi}{6} + 2 \pi k$ 2. $\arccos \left(- \frac{\sqrt{3}}{2}\right) + 2 \pi k = \pi - \arccos \left(\frac{\sqrt{3}}{2}\right) + 2 \pi k = \pi - \frac{\pi}{6} + 2 \pi k = \frac{5 \pi}{6} + \pi k$ $- \arccos \left(- \frac{\sqrt{3}}{2}\right) + 2 \pi k = \pi + \arccos \left(\frac{\sqrt{3}}{2}\right) + 2 \pi k = \pi + \frac{\pi}{6} + 2 \pi k = \frac{7 \pi}{6} + 2 \pi k$ Which solution is correct? My or here: https://socratic.org/questions/how-do-you-solve-cos-2x-3-4

You should consider two cases as follows

(1) $\setminus \cos x = \setminus \frac{\sqrt{3}}{2} = \setminus \cos \left(\setminus \frac{\pi}{6}\right)$

$x = 2 k \setminus \pi \setminus \pm \setminus \frac{\pi}{6}$

Where k is any integer

(2) $\setminus \cos x = - \setminus \frac{\sqrt{3}}{2} = \setminus \cos \left(\frac{5 \setminus \pi}{6}\right)$

$x = 2 k \setminus \pi \setminus \pm \frac{5 \setminus \pi}{6}$

Where k is any integer

Method-2 In general,

$\setminus {\cos}^{2} \setminus \theta = \setminus {\cos}^{2} \setminus \alpha$ has solutions

$\setminus \theta = n \setminus \pi \setminus \pm \setminus \alpha$

Where, n is any integer

Hence, the given trig. equation has following solutions

$\setminus {\cos}^{2} x = \frac{3}{4}$

$\setminus {\cos}^{2} x = {\left(\setminus \frac{\sqrt{3}}{2}\right)}^{2}$

$\setminus {\cos}^{2} x = {\left(\setminus \cos \left(\setminus \frac{\pi}{6}\right)\right)}^{2}$

$\setminus {\cos}^{2} x = \setminus {\cos}^{2} \left(\setminus \frac{\pi}{6}\right)$

$x = n \setminus \pi \setminus \pm \setminus \frac{\pi}{6}$

Where, $n$ is any integer i.e. $n = 0 , \setminus \pm 1 , \setminus \pm 2 , \setminus \pm 3 , \setminus \ldots$

Jul 4, 2018

color(blue)(x=2kpi+-pi/6 ,kinZZ or color(blue)(x=2kpi+-(5pi)/6 ,kinZZ

#### Explanation:

Here, ${\cos}^{2} x = \frac{3}{4} \implies \cos x = \pm \frac{\sqrt{3}}{2}$
$\text{If}$ $\cos x = a , | a | \le 1$ ,$\text{then the general solution of eqn. is :}$

color(blue)(x=2kpi+-alpha, kinZZandalpha=arc cos(a)
$\left(i\right) \cos x = \frac{\sqrt{3}}{2} > 0$
$\implies \alpha = a r c \cos \left(\frac{\sqrt{3}}{2}\right) = a r c \cos \left(\cos \left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6}$
color(blue)(x=2kpi+-pi/6 ,kinZZ
$\left(i\right) \cos x = - \frac{\sqrt{3}}{2} < 0$
$\implies \alpha = a r c \cos \left(- \frac{\sqrt{3}}{2}\right) = \pi - a r c \cos \left(\frac{\sqrt{3}}{2}\right)$=$\pi - \frac{\pi}{6}$=$\frac{5 \pi}{6}$
color(blue)(x=2kpi+-(5pi)/6 ,kinZZ
......................................................................................................
You have to think on red color.

$2. a r c \cos \left(- \frac{\sqrt{3}}{2}\right) + 2 \pi k$
=pi-arccos(sqrt3/2)+2pik=pi-pi/6+2pik=(5pi)/6+color(red)(pik

So, replace , color(red)(pik to2pik ,k inZZ

$\mathmr{and} a r c \cos \left(- \frac{\sqrt{3}}{2}\right) = \pi - a r c \cos \left(\frac{\sqrt{3}}{2}\right)$

$\implies \textcolor{red}{-} \left\{a r c \cos \left(- \frac{\sqrt{3}}{2}\right)\right\} = \textcolor{red}{-} \left\{\pi - a r c \cos \left(\frac{\sqrt{3}}{2}\right)\right\}$
=color(red)(-pi+arccos(sqrt3/2)=-pi+pi/6=(-6pi+pi)/6=-(5pi)/6

So, Replace , color(red)((7pi)/6 to-(5pi)/6

Also, remember to write : color(red)(kinZZto it is necessary.

Jul 5, 2018

$x = \frac{\pi}{6} \pm \pi n$
$x = \frac{5 \pi}{6} \pm \pi n$
Where $n$ is an element of all integers
n∈Z

#### Explanation:

An easy way to approach this would be to use the unit circle instead of using inverse trig:

$\cos x = \pm \frac{\sqrt{3}}{2}$

Now we know using the unit circle, the solutions are $\frac{\pi}{6} , \frac{5 \pi}{6} , \frac{7 \pi}{6} , \frac{11 \pi}{6}$

So the easiest way to write the solution set including all solutions would be:

$x = \frac{\pi}{6} \pm \pi n$
$x = \frac{5 \pi}{6} \pm \pi n$

Where $n$ is an element of all integers
n∈Z

graph{(cosx)^2-3/4 [-10, 10, -5, 5]}