Question

solve 8cos^2(x)-6cos(x)+1=0

Answer 1

that's the quadratic equation
let cos(x) = y to make it simple \[\huge\rm 8y^2-6y+1=0\]
now can you factor the quadratic equation ?

Answer 2

A)yes
B) no
C) maybe
pick one

Answer 3

yes

Answer 4

okay :=)) good
try to factor and then post your work

Answer 5

8y^2 -6y +1 = (2y-1)(4y-1) ---> 2y*4y=8y^2, 2y*-1=-2y, -1*4y= -4y, -1*-1=1 therefore 8y^2 -2y -4y +1= 8y^2 -6y +1 so it is factorable and above is it's factored form

Answer 6

that's correct! nice job
\[\rm \color{ReD}{(2y-1)}\color{blue}{(4y-1)}=0\]
set them equal to zero
\[\rm \color{reD}{2y-1}=0\] and \[\rm \color{blue}{4y-1}=0\]
solve for y

Answer 7

btw we should multiply the same two numbers to get the product of AC (a=leading coefficient c=constant term)so instead multiplying separately 2y and 4y by -1
you can just do -2y * -4y = 8y^2 \[\rm negative*negative =positive \]

Answer 8

alright I will keep that in mind thank you

Answer 9

bec \[\rm -2y*-4y=8y^2\]
\[\rm -2y-6y=-6y\]
\[2y+4y\cancel{= } -6y\]

Answer 10

but that's the same thing after we multiply them by negative one
alright


so what do you get for y ? :=))

Answer 11

y= 1/2 , 1/4

Answer 12

that's correct now we can replace y with cos(x)
so cos(x) =1/2 , cos(x)=1/4
to find an angle we should take invs cos
here is an example \[\large\rm \sin(x) = a \rightarrow x=\sin^{-1}(a)\]
invssin is same as arcsin
invscos is same as arccos

Answer 13

^^in that example x is an angle

Answer 14

so arccos(1/2) = X1, and arccos(1/4) = X2.

Answer 15

yes that's correct you can use unit circle for 1/2 but i guess you have to use calculator for 1/4
i hope you know on the unit circle x-coordinate = cos
and y-coordinate = sin

Answer 16

alirght so x = 1/3pi and x = 1.318...

Answer 17

oh do you need in radian or degree ?
if you need radian then keep it in fraction form

Answer 18

there is another angle where cos =1/2

Answer 19

don't worry it is all good, huge help btw thank you!

Answer 20

np :=))