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

1 Answer
Alan P.
May 21, 2016

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.

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