Odgovor:
# x = 2 / 3kpi + -pi / 9 # in # x = 2 / 3kpi + - (2pi) / 9 #
Pojasnilo:
Kot # | 2cos3x | = 1 #, imamo
prav tako # 2cos3x = 1 # t.j. # cos3x = 1/2 = cos (pi / 3) #
in # 3x = 2 kpi + -pi / 3 # ali # x = 2 / 3kpi + -pi / 9 #
ali # 2cos3x = -1 # t.j. # cos3x = -1 / 2 = cos ((2pi) / 3) #
in # 3x = 2 kpi + - (2pi) / 3 # ali # x = 2 / 3kpi + - (2pi) / 9 #
