Question

x=[0, 2pi] find the sum of the roots of the equation:

Answer 1

\[4\left( \cos x \right)^{3}-3\cos x=0\]

Answer 2

well factor out cos(x) and you have

\[\cos(x)(4\cos^2(x) - 3) = 0\]

so you now need to factor the quadratic

and you get

\[\cos(x)(2\cos(x) - \sqrt{3})(2\cos(x) + \sqrt{3}) = 0\]

I'll leave you to find the values of x in the given domain that make the equation true

Answer 3

okay

Answer 4

do you know what to do next...?

Answer 5

I think i have to find 2cos(x)=sqrt(3) ?

Answer 6

4(cosx)^3=3(cosx)
4(cosx)^2=3
(Cosx)^2=(3/4)
Cosx=\[\pm \sqrt{3/4}\]
Cos x = \[\frac{ \sqrt{3} }{2 }\]

Answer 7

lol... there you go.... why worry about learning where people just post solutions...

but there is a missing solution...

Answer 8

I found this sqrt(3)/2 but wasn't sure what that meant

Answer 9

well thats correct as posted by @aj.est1979

there is a solution where

\[\cos(x) = \pm \frac{\sqrt{3}}{2}\]

there are exact value angles... so easy to find....

go back to my factorization to find the other value cos(x) can take

Answer 10

but remember... you need angles...

Answer 11

cos(x)=0, x=pi/2?

Answer 12

really helpful

Answer 13

that correct @lucaz it has cos(x) = 0

and in the given domain, cos(x) = 0 for 2 angles...

one is \[\frac{\pi}{2}\]

and there is another...

Answer 14

I think it's 3pi/2

Answer 15

thats correct...


so now you have all the solutions....

Answer 16

hey, this kind of problem the domain is always ginen with pi, or it could be in degrees?

Answer 17

thats correct... your answers are expressed as radians.... if the domain was degrees just change them to radians..

pi/2 = 90 3pi/2 = 270

Answer 18

right, thank you