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Find the sum of all values of \(\theta\) such that \(\theta\in[0,2\pi]\) and \(\cos(\theta)+4\sec(\theta)=8 \)
 

 Oct 29, 2019
 #1
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i'm using x in place of theta

 

cos (x)  + 4 sec(x)  =  8

 

cos (x)  + 4 / cos(x)  = 8         multiply through by cos (x)

 

cos^2x  + 4  = 8cos(x)

 

cos^2 (x) - 8cos(x) + 4  =  0

 

Let cos(x)  = m       and we have that

 

m^2 - 8m  + 4  = 0

 

m^2  - 8m  =   - 4      complete the square on m

 

m^2  - 8m + 16  =  -4 + 16

 

(m - 4)^2  =  12        take both roots

 

m - 4  =  ±2√3

 

m =  ±2√3 + 4

 

Only the negative root makes sense

 

So

 

m  =  cos x  =  (-2√3  + 4)

 

Take the cosine inverse

 

arcos ( -2√3 + 4)  =  x  ≈  1.005 rads     and  (2pi - 1.005)  ≈ 5.28 rads  

 

So....the sum of these  ≈  6.285 rads

 

cool cool cool

 Oct 29, 2019
edited by CPhill  Oct 29, 2019

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