Question

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

`cos^2x=3/4`
This is already covered here:
https://socratic.org/questions/how-do-you-solve-cos-2x-3-4
But my solution:
1. `cosx=(sqrt3)/2`
2. `cosx=-(sqrt3)/2`
So:
1. `arccos((sqrt3)/2)+2pik=pi/6+2pik`
`-arccos((sqrt3)/2)+2pik=-pi/6+2pik`
2. `arccos(-(sqrt3)/2)+2pik=pi-arccos((sqrt3)/2)+2pik= pi-pi/6+2pik=(5pi)/6+pik`
`-arccos(-(sqrt3)/2)+2pik=pi+arccos((sqrt3)/2)+2pik=pi+pi/6+2pik=(7pi)/6+2pik`
Which solution is correct? My or here:
https://socratic.org/questions/how-do-you-solve-cos-2x-3-4

Answer 1

You should consider two cases as follows

(1) `\cos x=\sqrt3/2=\cos(\pi/6)`

`x=2k\pi\pm\pi/6`

Where k is any integer

(2) `\cos x=-\sqrt3/2=\cos({5\pi}/6)`

`x=2k\pi\pm{5\pi}/6`

Where k is any integer

Method-2 In general,

`\cos^2\theta=\cos^2\alpha` has solutions

`\theta=n\pi\pm\alpha`

Where, n is any integer

Hence, the given trig. equation has following solutions

`\cos^2x=3/4`

`\cos^2x=(\sqrt3/2)^2`

`\cos^2x=(\cos(\pi/6))^2`

`\cos^2x=\cos^2(\pi/6)`

`x=n\pi\pm \pi/6`

Where, `n` is any integer i.e. `n=0, \pm1, \pm2, \pm3, \ldots`

Answer 2

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

Explanation:

Here, `cos^2x=3/4=>cosx=+-sqrt3/2`
`"If"` `cosx=a, |a|<=1` ,`"then the general solution of eqn. is :"`

`color(blue)(x=2kpi+-alpha, kinZZandalpha=arc cos(a)`
`(i)cosx=sqrt3/2 >0`
`=>alpha=arc cos(sqrt3/2)=arc cos(cos(pi/6))=pi/6`
`color(blue)(x=2kpi+-pi/6 ,kinZZ`
`(i)cosx=-sqrt3/2 < 0`
`=>alpha=arc cos(-sqrt3/2)=pi-arc cos(sqrt3/2)`=`pi-pi/6`=`(5pi)/6`
`color(blue)(x=2kpi+-(5pi)/6 ,kinZZ`
......................................................................................................
You have to think on red color.

Compare your answer with red color to correct your answer.

`2. arc cos(-sqrt3/2)+2pik`
`=pi-arccos(sqrt3/2)+2pik=pi-pi/6+2pik=(5pi)/6+color(red)(pik`

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

`and arc cos(-sqrt3/2)=pi-arc cos(sqrt3/2)`

`=>color(red)(-){arc cos(-sqrt3/2)}=color(red)(-){pi-arc cos(sqrt3/2)}`
`=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.

Answer 3

`x=pi/6+-pin`
`x=(5pi)/6+-pin`
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:

`cosx=+-sqrt3/2`

Now we know using the unit circle, the solutions are `pi/6, (5pi)/6, (7pi)/6, (11pi)/6`

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

`x=pi/6+-pin`
`x=(5pi)/6+-pin`

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

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