Question #00b38

2 Answers
Apr 13, 2016

cos 3x=1/2
3x=cos^-1(1/2)
3x=+-pi/3+2pi n
x=+-pi/9 + 2/3 pi n

Apr 14, 2016

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

Explanation:

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^@