How do you solve #cos2x=(-5/12) #?

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Answer 1 · 2016-10-29T04:51:36

Let #cos alpha =-5/12=>alpha =cos^-1(-5/12)#

Given equation #cos2x=-5/12# becomes

#cos2x=cosalpha#

#=>2x=2npi+-alpha,where " " n in ZZ#

So #x=npi+-alpha/2#

By using calculator we have

#alpha =cos^-1(-5/12)=0.6368pi#
And
#x=npi+-(0.6368pi)/2=np+-0.3184pi#