How do you solve #3cos^2x-6cosx=sin^2x-3#?

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Answer 1 · 2016-11-26T14:29:41

The solutions are #S= {2kpi,pi/3+2kpi,-pi/3+2kpi}#, #k in ZZ#

We use

#cos^2x+sin^2x=1#

#3cos^2x-6cosx=sin^2x-3#

#3cos^2x-6cosx=1-cos^2x-3=-cos^2x-2#

#4cos^2x-6cosx+2=0#

Dividing by #2#

#2cos^2x-3cosx+1=0#

This is a quadratic equation

The discriminant is

#Delta=b^2-4ac=9-4*2*1=1#

#Delta>0#, we have2 real roots

#cosx=(-b+-sqrt(Delta))/(2a)#

#cosx=(3+-1)/4#

#cosx=1# or #cosx=1/2#

#x=0+2kpi# or #x=pi/3+2kpi# or #x=-pi/3+2kpi#

#k in ZZ#