How do you solve #2 cos ^2 x - 5 cosx + 2 = 0#?

1 Answer
Alan P.
Mar 24, 2016

#x=+-pi/3 (+2npi, nin ZZ)#

Explanation:

Given
#color(white)("XXX")2cos^2x-5cos x + 2 =0#

Let #c=cos(x)#
then
#color(white)("XXX")2c^2-5c+2=0#

Factoring:
#color(white)("XXX")(c-2)(2c-1)=0#

Which implies
#color(white)("XXX")c=2#
or
#color(white)("XXX")c=1/2#

Since #c=cos(x)# and #cos(x)# has a Range of #[-1,+2]#
#color(white)("XXX")c=2# is an extraneous result

Therefore
#color(white)("XXX")c=cos(x)=1/2#
which is one of the standard angles with
#color(white)("XXX")x=+-pi/3# when #x in [0,2pi)#
or, in general
#color(white)("XXX")x=+-pi/3 +2npi, nin ZZ#

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