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^@