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?

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.

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

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.

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.

Hmmm. :mad: What answer was said to be the correct one?

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:

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. :(

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...

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