How do you solve #2cos^2x-2sin^2x=1# and find all solutions in the interval #0<=x<360#?

1 Answer
maganbhai P.
Mar 14, 2018

#x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6#
Where, #0<=x<360^0=>x in[0,2pi)#

Explanation:

#2cos^2x-2sin^2x=1=>cos^2x-sin^2x=1/2#
#=>cos2x=1/2=>cos2x=cos(pi/3)#
#=>2x=2kpi+-pi/3,kinZ#
#color(red)(=>x=kpi+-pi/6,kinZ)#
#kinZ=>k=0,+-1.+-2,+-3,..etc#
#k=0=>color(red)(x=pi/6in[0,2pi))#
#k=1=>color(red)(x=(5pi)/6in[0,2pi)orx=(7pi)/6in[0,2pi)#
#k=2=>color(red)(x=(11pi)/6in[0,2pi))orx=(13pi)/6!in[0,2pi)#...
#AAkinZ,k>3=>x!in[0,2pi)#
#k=-1=>x=(-5pi)/6!in[0,2pi)orx=(-7pi)/6!in[0,2pi)#..
#AAkinZ,k<-1=>x!in[0,2pi)#
Hence, #x=pi/6,(5pi)/6,(7pi)/6,(11pi)/6#

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