We can first divide both side of equation by #2# to eliminate the value of #2# on #2cos3x#;
#(2cos3x)/2=1/2#
#cos3x=1/2 #
The fact that the trigonometry equation of #cos3x# has positive value means that it is located in Quadrant 1 and Quadrant 4
#x# initially is in the range from #0# until #2pi# or #360# and expressed as;
#x in [0^@,360^@]#
But since #x# now is in the form of #3x# we must multiply all the equation in #x in [0,360^@]# with #3# and we get;
#3x in [0^@,1080^@]#
The trigonometry equation now gives us a range of
Quadrant 1, 4, 5, 8, 9, and 12 .
Multiply both side in #cos3x=1/2# with #cos^-1# to cancel out #cos# in #cos3x#;
#cancel(cos^-1) cancel(cos)3x=cos^-1 (1/2)#
#3x=60^@#
And since #3x# is present in Quadrant 1, 4, 5, 8, 9, and 12 ;
#3x=60^@,(360^@-60^@),(360^@+60^@),(720^@-60^@),(720^@+60^@),(1080^@-60^@)#
#3x=60^@,300^@,420^@,660^@,780^@,1020^@#
Divide both side by #3# and we get the very final answer of all values of #x#;
#(3x)/3=60^@/3,300^@/3,420^@/3,660^@/3,780^@/3,1020^@/3#
#x=20^@,100^@,140^@,220^@,260^@,340^@#
