Sagot:
# x = 2 / 3kpi + -pi / 9 # at # x = 2 / 3kpi + - (2pi) / 9 #
Paliwanag:
Bilang # | 2cos3x | = 1 #, meron kami
alinman # 2cos3x = 1 # i.e. # cos3x = 1/2 = cos (pi / 3) #
at # 3x = 2kpi + -pi / 3 # o # x = 2 / 3kpi + -pi / 9 #
o # 2cos3x = -1 # i.e. # cos3x = -1 / 2 = cos ((2pi) / 3) #
at # 3x = 2kpi + - (2pi) / 3 # o # x = 2 / 3kpi + - (2pi) / 9 #
