How can this be sovled?

#2sin^2x = 2 + cosx#

2 Answers
Feb 25, 2018

#x=pi/2, (3pi)/2# or #x=(2pi)/3 or (4pi)/3#

Explanation:

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

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

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

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

#cosx=0#
#cosx=-1/2#

#x=pi/2, (3pi)/2# or #x=(2pi)/3 or (4pi)/3#

Feb 25, 2018

#x in {(pi/2)+kpi, (pi+-pi/3)+k * 2pi}, k in ZZ#

Explanation:

If #2sin^2(x)=2+cos(x)#
then
#color(white)("XX")2(1-cos^2(x))=2+cos(x)#

#color(white)("XX")2-2cos^2(x) = 2+cos(x)#

#color(white)("XX")2cos^2(x)+cos(x)=0#

by factoring:
#color(white)("XX"){: (,cos(x)=0,color(white)("xx")"or"color(white)("xx"),2cos(x)=-1), ("for "x in [0,2pi],,,), (,x= pi/2" or " (3pi)/2,,cos(x)=-1/2), (,,,x=(2pi)/3" or "(4pi)/3) :}#