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Answer 1 · 2017-12-15T19:45:11

#=> x = (2npi)/3 pm pi/9 + pi/18 #

# n in ZZ #

Rearanging:

#=> 2cos(3x - pi/6) = 1#

#=> cos(3x - pi/6) = 1/2 #

Now considering how #cos(pi/3) = 1/2 #

#=> cos(3x - pi/6) = cos(pi/3) #

Now considering the general solution of #cosx#

if #cosx = cos gamma #

#=> x = 2pin pm gamma #

Hence:

#=> 3x- pi/6 = 2pin pm pi/3 #

Rearanging:

#=> 3x = 2pin pm pi/3 +pi/6 #

#=> x = (2npi)/3 pm pi/9 + pi/18 #

# n in ZZ #