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How do you solve #2cos^2x+3cosx-2=0#?

1 Answer

Nov 27, 2016

Let #n=cosx#

#2n^2+3n-2=0#

Solve for n by factoring:
#2n^2-n+4n-2=0#
#n(2n-1)+2(2n-1)=0#
#(n+2)(2n-1)=0#

#n+2=0#
#n=-2#

#2n-1=0#
#n=1/2#

Substitute #n=cosx# in for those two answers:
#cosxcancel(=)-2#
Cosine can never equal -2.

#cosx=1/2#

#x=pi/3, (5pi)/3#

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