Question #00b38

2 Answers
Apr 13, 2016

cos 3x=1/2cos3x=12
3x=cos^-1(1/2)3x=cos1(12)
3x=+-pi/3+2pi n3x=±π3+2πn
x=+-pi/9 + 2/3 pi nx=±π9+23πn

Apr 14, 2016

x=20^@,100^@,140^@,220^@,260^@,340^@x=20,100,140,220,260,340

Explanation:

We can first divide both side of equation by 22 to eliminate the value of 22 on 2cos3x2cos3x;

(2cos3x)/2=1/22cos3x2=12

cos3x=1/2 cos3x=12

The fact that the trigonometry equation of cos3xcos3x has positive value means that it is located in Quadrant 1 and Quadrant 4

xx initially is in the range from 00 until 2pi2π or 360360 and expressed as;

x in [0^@,360^@]x[0,360]

But since xx now is in the form of 3x3x we must multiply all the equation in x in [0,360^@]x[0,360] with 33 and we get;

3x in [0^@,1080^@]3x[0,1080]

The trigonometry equation now gives us a range of
Quadrant 1, 4, 5, 8, 9, and 12 .

Multiply both side in cos3x=1/2cos3x=12 with cos^-1cos1 to cancel out coscos in cos3xcos3x;

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^@