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Solve the equation 2 cos x - sec x = 1 on the interval [0, 2π). how would i do this

Guest Mar 25, 2015

Best Answer 

 #1
avatar+87301 
+5

2 cos x - sec x = 1   .....note sec x = 1/cosx  so we have

2cosx - 1/cosx  = 1      multiply through by cos x

2cos^2 x  - 1  =  cos x    rearrange

2cos^2 x - cos x - 1  = 0     factor

(2 cos x + 1)(cos x - 1) = 0      so either

2 cos x + 1   = 0

2 cos x = -1

cos x = -1//2     and this occurs at  2pi/3 and 4pi/3

OR

cosx - 1  = 0   

cos x =  1    and this occurs at  0

So the solutions on the given interval are  0, 2pi/3 and 4pi/3

 

  

CPhill  Mar 25, 2015
 #1
avatar+87301 
+5
Best Answer

2 cos x - sec x = 1   .....note sec x = 1/cosx  so we have

2cosx - 1/cosx  = 1      multiply through by cos x

2cos^2 x  - 1  =  cos x    rearrange

2cos^2 x - cos x - 1  = 0     factor

(2 cos x + 1)(cos x - 1) = 0      so either

2 cos x + 1   = 0

2 cos x = -1

cos x = -1//2     and this occurs at  2pi/3 and 4pi/3

OR

cosx - 1  = 0   

cos x =  1    and this occurs at  0

So the solutions on the given interval are  0, 2pi/3 and 4pi/3

 

  

CPhill  Mar 25, 2015

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