View Full Version : Trigonometric Equation
kayladawn
Apr 28, 2009, 12:51 PM
I am having trouble with the following trigonometry problem:
cos2x= -1/2
I tried to turn cos2x into 2cos^2x-1 according to the double angle identity, but then I was unable to factor the problem. Where have I gone wrong, or where do I go from here?
Perito
Apr 28, 2009, 02:13 PM
I'm not sure why you need the identity. Can't you solve it like this or isn't it allowed?
cos(2x) = -\frac 12
2x = cos^{-1}(-\frac 12) = \pm\,120 \, degrees
x = \pm\, 60 \, degrees
I guess you could do it this way:
cos(2x) = 2cos^2(x) - 1 = -\frac 12
cos^2(x) = 0.25
cos(x) = \sqrt{0.25} = \pm \frac 12
And now you have two roots and you still have to take the ArcCos, so you haven't gained much.
Unknown008
Apr 29, 2009, 08:16 AM
My math tell me that
cos(2x) = -\frac{1}{2}
x= 60 and 120 degrees... for 0 < x < 360
BUT if you do through identity, x = 30, 60, 120 and 150 for 0 < x < 360
Perito
Apr 29, 2009, 08:24 AM
My math tell me that
cos(2x) = -\frac{1}{2}
x= 60 and 120 degrees... for 0 < x < 360
and if you do through identity, x = 30, 60, 120 and 150 for 0 < x < 360
I'm confused. Cos(60) = 0.5, not -0.5. You need to be in the 2nd or 3rd quadrant for it to be -0.5. So 2x = ±120 (or 240), x=±60 on 0 < x < 360. Oh. I guess that does give x=120 also. My bad. 30 degrees and 150 degrees certainly don't fit the original problem, however.
kayladawn
Apr 29, 2009, 09:31 AM
Perito, when I turned in my homework, I found cos x= 1/2 without the identity as you worked it and then I found my solution set, but I got it wrong.
Perito
Apr 29, 2009, 10:05 AM
Hmmm. :mad: What answer was said to be the correct one?
Unknown008
Apr 29, 2009, 10:13 AM
I'm confused. cos(60) = 0.5, not -0.5. You need to be in the 2nd or 3rd quadrant for it to be -0.5. So 2x = ±120 (or 240), x=±60 on 0 < x < 360. Oh. I guess that does give x=120 also. My bad. 30 degrees and 150 degrees certainly don't fit the original problem, however.
Yes,
cos(60)=0.5
but
cos(2x60)=-0.5
From cos(2x)=-0.5
And the plus/minus sign for cos(2x)=±0.5 (for the cos^2(x) = 0.25)
will give the roots in all four quadrants, so
2x = 60, 120, 240, 300
x = 30, 60, 120, 150
:confused:
Perito
Apr 29, 2009, 10:17 AM
And the plus/minus sign for cos(2x)=±0.5 (for the cos^2(x) = 0.25)
will give the roots in all four quadrants, so
2x = 60, 120, 240, 300
x = 30, 60, 120, 150
I'm having a tough time with the ±0.5. The original problem only said -0.5. If you go through the trig identity, you get ±0.5, but +5 doesn't fit the original problem so you have to discard it. Let me know where I'm going wrong. :(
Unknown008
Apr 29, 2009, 10:40 AM
Yeah, I know that those do not fit in the original question... :o was just posting just in case there was a need for passing through trig identity.
Ok, edited my first post in the thread...
Unknown008
Apr 29, 2009, 11:28 AM
Perito agrees: Good. I wonder why they marked her wrong.
Or was it in rads? Then the answers would be
\frac{\pi}{3}, \frac{2\pi}{3}
or
1.05 , 2.09