How do you solve for the exact solutions in the interval [0,2pi] of #cos3x=(1/2)#?

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Answer 1 · 2016-04-18T21:12:20

#S={pi/9, (5pi)/9,(7pi)/9, (11pi)/9, (13pi)/9, (17pi)/9 }#

#3x=cos^-1 (1/2)#

#3x=+- pi/3 +2pin#

#x=+- pi/3 xx 1/3+2pin xx 1/3#

#x=+-pi/9+(2pi)/3 n#

Now pick n values (positive and negative integers) into the equation

#n=0, x=pi/9,-pi/9#

#n=1, x=(5pi)/9,(7pi)/9#

#n=2, x=(11pi)/9, (13pi)/9#

#n=3, x=(17pi)/9,(19pi)/9#

#S={pi/9, (5pi)/9,(7pi)/9, (11pi)/9, (13pi)/9, (17pi)/9 }#