How do you solve #cos2x=[sqrt(2)/2]# over the interval 0 to 2pi?

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Answer 1 · 2016-04-17T20:37:57

#S={pi/8, (7pi)/8, (9pi)/8,(15pi)/8}#

#2x=cos^-1 (sqrt 2/2)#

#2x=+-pi/4 + 2pin#

#x=+- pi/8 + pi n#

#n=0, x=pi/8, -pi/8#

#n=1, x=(9pi)/8, (7pi)/8#

#n=2, x=(17pi)/8, (15pi)/8 #

#S={pi/8,(7pi)/8, (9pi)/8,(15pi)/8}#