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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