Author 
Topic 

jfg4707
Average Member
USA
19 Posts 
Posted  10/05/2007 : 21:33:32

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=1.3693=1.125 The reference angle of 1.125 is 1.125. So, x=1.125 and 21.125=5.158.
The correct answers are 1/914 amd 4.653
Thank you for any help you can give me. 


tkhunny
Advanced Member
USA
1001 Posts 
Posted  10/05/2007 : 21:49:25

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 


uberclay
Advanced Member
Canada
159 Posts 
Posted  10/06/2007 : 00:03:52

quote: Originally posted by tkhunny
1.369  3 = 1.631 NOT 1.125



jfg4707
Average Member
USA
19 Posts 
Posted  10/06/2007 : 01:15:52

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 


skeeter
Advanced Member
USA
5634 Posts 
Posted  10/06/2007 : 09:01:03

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 


jfg4707
Average Member
USA
19 Posts 
Posted  10/06/2007 : 13:27:54

Skeeter, thank you very much. 


Cortez Mos
Junior Member
USA
3 Posts 
Posted  10/17/2007 : 23:21:20

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


the_hill1962
Advanced Member
USA
1469 Posts 


Topic 
