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^@#