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avatar+784 

Find the intersection between graph r=1 and r=2cosθ

i did cosθ= 1/2

cos^-1 1/2 = 1/3π (radiance)

from here on ward i was told to add something to this which should give me 5/3π

and thus my 2 intersection should be 

(1, 1/3π) (1, 5/3π) 

please tell me what i do in order to get to 5/3π

 Feb 27, 2019
 #1
avatar+98168 
+2

Set the r's equal  and we have that

 

2cos θ = 1       divide both sides by 2

 

cos  θ  =  1/2

 

arccos (1/2) =  θ

 

Remember that arccos  has a range of  [-pi,2, pi/2 ]

 

Note that this happens   at  pi/3    and at  -pi/3 =  5pi/3       if we are on the interval [0, 2pi ]

 

So...the intersection points   are

 

  (r,  θ) =   (  1, pi/3)   and (1, 5pi/3 )      

 

 

cool cool cool

 Feb 27, 2019
 #2
avatar+784 
+1

can you please explain how you find out -pi/3 = 5pi/3 

YEEEEEET  Feb 27, 2019
 #3
avatar+98168 
+2

Notice that -pi/3  =  -60°   =   360 - 60 =    300°  =    5pi / 3

 

 

cool cool cool

 Feb 27, 2019
 #4
avatar+784 
+1

ah ok thank you 

one more question: does this method apply to all polar graphs such as when you have to solve tan and sin or is there a different method to do so

YEEEEEET  Feb 27, 2019
edited by YEEEEEET  Feb 27, 2019
 #5
avatar+98168 
+1

I don't know the answer to that YEEEEEET.....I suppose it depends upon what we are given

 

We can always set the r's equal......but.....the resulting equation may (or may not) be easy to solve algebraically

 

 

cool cool cool

CPhill  Feb 27, 2019
 #6
avatar+784 
+1

ok thank you so much for helping!

YEEEEEET  Feb 27, 2019

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