How do you find all values of x in the interval [0, 2pi] in the equation #cos(2x)+cos(x)=2#?

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Answer 1 · 2016-04-18T21:49:18

#S=2pin#

Use the Property: #cos2A=2cos^2A-1#

#2cos^2x-1+cosx=2#

#2cos^2x+cosx-3=0#

#(2cosx+3)(cosx-1)=0#

#cosx=-3/2 or cosx=1#

#x=cos^-1(-3/2) or x=cos^-1 1#

#x=#undefined, #x=2pin#

#S=2pin#