Question

How do you solve `8cos^2 x - 2cos x=1` for x in the interval [0,2pi)?

Answer 1

`x in {pi/3,(5pi)/3,0.58pi,1.42pi}`

Explanation:

Given
`color(white)("XXX")8cos^2(x)-2cos(x)=1`

Substituting `t` for `cos(x)`
`color(white)("XXX")8t^2-2t=1`

`color(white)("XXX")8t^2-2t-1=0`

`color(white)("XXX")(2t-1)(4t+1)=0`

#{: ("either",,"or ",), (,t=1/2,,t=-1/4), (,rarrcos(x)=1/2,,rarrcos(x)=-1/4), (,rarr x=+-arccos(1/2),,rarrx=pi+-arccos(1/4)), (,rarr x= +-pi/3,,rarr x=(1+-0.419569)pi) :}#

Note 1:
The calculation for `arccos(1/2)` gives a value in the range `(-pi,+pi]`
The negative value, `-pi/3`, was converted by the addition of `2pi` (one full rotation) to give `(5pi)/3`

Note 2:
The value for `arccos(1/4)` was obtained using a calculator and is only approximate. I further approximated it to the values shown in the "Answer" by rounding to two digits to the right of the decimal point.