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jfg4707
Average Member

USA
19 Posts

Posted - 10/05/2007 :  21:33:32  Show Profile
Hi,

I am trying to solve a trigonometry equation, but I keep getting the wrong answer. I can't figure out what I am doing wrong.

Problem

Find all solution for the following equation on the interval 0x2: 5cos(x+3)=1.

Solution
5cos(x+3)=1
cos(x+3)=1/5
x+3=arccos(1/5)=1.369
x=1.369-3=-1.125
The reference angle of -1.125 is 1.125. So,
x=1.125 and 2-1.125=5.158.

The correct answers are 1/914 amd 4.653

Thank you for any help you can give me.
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tkhunny
Advanced Member

USA
1001 Posts

Posted - 10/05/2007 :  21:49:25  Show Profile
1.369 - 3 = -1.631 NOT -1.125

-1.631 + 2 = 4.653

You find the other one.

(Typo Repaired - Thanks uberclay. How did I get 4.653 right?)

Edited by - tkhunny on 10/06/2007 01:14:39
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uberclay
Advanced Member

Canada
159 Posts

Posted - 10/06/2007 :  00:03:52  Show Profile
quote:
Originally posted by tkhunny

1.369 - 3 = -1.631 NOT -1.125

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jfg4707
Average Member

USA
19 Posts

Posted - 10/06/2007 :  01:15:52  Show Profile
tkhunny,

With x= 4.653, a third quadrant angle, I assumed that the second angle must be in Quadrant 2, so they would have the same sign when cosine was applied; and it is in Quadrant II, i.e., x= 1.914. But I have not been able to derive 1.914.

I thought that finding the reference angle of 4.653 and then stubracting it from would give me the correct answer, but it does not: 4.653-=1.511 and -1.511=1.631. Everything else I have tried fails to produce 1.914. Can you explain to me what I shoulld be doing?

Thank you
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skeeter
Advanced Member

USA
5634 Posts

Posted - 10/06/2007 :  09:01:03  Show Profile
arccos(1/5) = 1.369, a quad I angle.
x = 1.3694 - 3 = -1.631

arccos(1/5) = 2 - arccos(1/5) = 4.914, a quad IV angle.
x = 4.914 - 3 = 1.914
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jfg4707
Average Member

USA
19 Posts

Posted - 10/06/2007 :  13:27:54  Show Profile
Skeeter, thank you very much.
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Cortez Mos
Junior Member

USA
3 Posts

Posted - 10/17/2007 :  23:21:20  Show Profile
im trying to solve
Cos (-10) cos 35 + sin (-10) sin 35 each number in degrees please help me out here!
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the_hill1962
Advanced Member

USA
1468 Posts

Posted - 10/18/2007 :  06:34:57  Show Profile
This new problem was previously posted at
http://www.mathgoodies.com/forums/topic.asp?TOPIC_ID=32967
quote:
Originally posted by Cortez Mos

im trying to solve
Cos (-10) cos 35 + sin (-10) sin 35 each number in degrees please help me out here!

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