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Post by elocina on May 17, 2006 23:21:04 GMT -5
I'm just wondering if there's anyone who can help me with this:
2sin(squared)X + 7cosX + 2 = 0
or
(Square root of three) times cot3X + 1 = 0
If I can't be helped by next week, that's alright. I've got my final on Wednesday.
Thanx,
C
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Post by Snoink on May 17, 2006 23:31:14 GMT -5
What do the instructions say for these problems?
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Post by elocina on May 18, 2006 16:43:40 GMT -5
Nothing, but I assume they want the answer in degrees.
I THINK the first one is supposed to be factorable.... The second one, I asked a friend. How the second one goes is: (Square root of three) times cot3X + 1 = 0 cot3X=-1/(square root of three) tan3x=-(square root of three) ((tanx=sinx/cosx)) so... sinx=(1/2) cosx=(1/(square root of three)) with either vairable negative.
On the unit circle, it equals 120 degrees, and 330 degrees; but those have to be divided by three since those angles equal tanx not tan3x...
Making any sense? The end answer was 40 and 110 degrees on the second one.
First one I can't quite figure out, but it's similar.
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Post by Meshugenah on May 18, 2006 23:25:21 GMT -5
the first one? Yeah, I think it's factorable..
lemme see.. nope! not factorable. Ok, I think I got it.
you have 2sin^2x, so take that and make 2(1 - cos^2x) + 7cosx +2 = 0. multiply out to get 2 - 2cos^2x +7cosx +2 = 0 Condense: -2cos^2x + 7cosx + 4 = 0 now, take out the -1, so: -(2cos^2x -7cosx -4) = 0.
factor to get (using this technique my teacher showed me, so use -(2x^2 - 7x -4) = 0): -(2x + 1)(x - 4). Set each factor to equal 0, so 2x + 1 = 0, and x - 4 = 0. SO, x = -1/2 or 4. arccos(1/2) = 30 degrees, I believe (I always mix 30 and 60, so don't take me word on that one), and teh arccos (4). Your answer will be in degress, and for the negative one, -30 (if I'm right), so you'll want to fix that to 330 or 210, I believe, still going with the 30 being right..
Hope that makes some sense, C!
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Post by elocina on May 19, 2006 10:01:44 GMT -5
OK, yes that makes sense.... answer should be 120, 240. (Only reason I know the answer is I looked in the back of the book. But the answer won't help me on the final.....) I think that's the right way to get the answer, though, Mesh. Thanks
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