How do you find the general solutions for #2 cos^2 x - cos x = 0#?

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Answer 1 · 2018-04-08T03:58:59

#x=pi/2+npi, x=pi/3+2npi, x=(5pi)/3+2npi#

All terms share the cosine in common. We can thus factor out a cosine:

#cosx(2cosx-1)=0#

We may now solve the following two:

a. #cosx=0#

#x=pi/2+npi#

As cosine is zero for #pi/2, (3pi)/2, (5pi)/2,...,pi/2+npi#
#2cosx-1=0#

#2cosx=1#

#cosx=1/2#

#x=pi/3+2npi#

#x=(5pi)/3+2npi#

As cosine is equal to (positive) one half in quadrants I, IV.