How do you solve #cos^2x= cosx# for x in the interval [0,2pi)?

1 Answer
sente
Mar 1, 2016

Move all terms to one side and factor the equation to find all possible values for #cos(x)# to find that the solution set to #cos^2(x) = cos(x)# restricted to #[0, 2pi)# is

#{0, pi/2, (3pi)/2}#

Explanation:

#cos^2(x) = cos(x)#

#=> cos^2(x) - cos(x) = 0#

#=> cos(x)(cos(x)-1) = 0#

#=> cos(x) = 0# or #cos(x) = 1#

On the interval #[0, 2pi)# we have

#cos(x) = 0 <=> x in {pi/2, (3pi)/2}#

and

#cos(x) = 1 <=> x = 0#

Thus our complete solution set is #x in {0, pi/2, (3pi)/2}#

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