# Trig problem (cos)

• Dec 4th 2005, 08:47 PM
killasnake
Trig problem (cos)
Hi I really don't know to do this problem.

The equation cos(3x) = 1/2

has two solutions between 0 and 120 degrees. The smaller is _____ degrees and the larger is _______ degrees.

Am I suppose to take cos(pi/3) / 3 and thats my answer?
• Dec 5th 2005, 03:40 AM
ticbol
Quote:

Originally Posted by killasnake
Hi I really don't know to do this problem.

The equation cos(3x) = 1/2

has two solutions between 0 and 120 degrees. The smaller is _____ degrees and the larger is _______ degrees.

Am I suppose to take cos(pi/3) / 3 and thats my answer?

No, you are not supposed to do that.
Instead, you are supposed to get the the arccosine of (1/2), and equate those to 3x to find x.

cos(3x) = 1/2
That means the cosine of an angle, here it is 3x, is positive 1/2.
In the 4 quadrants, where is cosine value positive?
Cosine is positive where the x-coordinate is positive. The x-coordinate is positive to the right of the y-axis, or at the 1st and 4th quadrants. Hence, the angle 3x is in the 1st or 4th quadrant.

cos(3x) = 1/2
3x = arccos(1/2)
3x = 60 degrees in the 1st quadrant.
3x = 360 -60 = 300 degrees in the 4th quadrant.

Therefore, x = 60/3 = 20 degrees.
or, x = 300/3 = 100 degrees.
• Dec 5th 2005, 03:02 PM
killasnake
oh! Thank you for the explanation.