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Find the number of solutions to \(\sec \theta + \csc \theta = \sqrt{15}\) where \(0 \le \theta \le 2 \pi\)

 Sep 27, 2019
 #1
avatar+104962 
+1

 

 

[ I'm using "A" for theta ]

 

sec A  + csc A  =   sqrt (15) 

 

1/cosA  + 1/sinA  = sqrt (15)

 

[sin A + cosA ]

___________  =    sqrt (15)

  sin A cos A     

 

sinA + cosA  = sqrt (15) * sinAcosA         square both sides

 

sIn^2A + 2sinAcosA + cos^2A   = 15sin^2Acos^2A 

 

1 + 2sinAcosA  =  15sin^2Acos^2A             rearrange as

 

15sin^2Acos^2A - 2sinAcosA - 1   =   0      factor as

 

(5sinAcosA + 1) (3sinAcosA - 1)  =    0

 

We have either that

 

5sinAcosA  + 1  = 0

5sinAcosA = -1

sinAcosA = -1/5

(1/2)sin(2A)  = -1/5

sin(2A)  =   -2/5

Let 2A  = M

sin (M)  =  -2/5

Using the sine inverse

arcsin (-2/5)  = M      ≈ ( - .412  + 2pi)rads            or       M =    (.412 + pi) rads

So  2A  =( -.412 + 2pi) rads                                            2A  =  (.412 + pi)rads

A  = -.206 + pi rads ≈ 2.94 rads                                        A  = .206 + pi/2 rads ≈ 1.77 rads

 

And M = (-.412 + 4pi)rads                                    or     M  =  (.412 + 3 pi)rads

2A  =  (-.412 + 4pi)rads                                                2A  = (.412 + 3pi) rads

A  =  -..206 + 2pi rads ≈ 6.07 rads                               A  =  .206 + (3/2)pi rads ≈  4.91 rads

 

 

Or

3sinAcosA  -  1  = 0

3sinAcosA = 1

sinAcosA  = 1/3

(1/2)sin (2A)= 1/3

sin(2A)  = 2/3

Let 2a  = M

sin(M) = 2/3

arcsin(2/3)  = M  ≈ .73 rads  

So

2A  = .73 rads              or        2A  = .(pi -.73) rads

A  ≈   .365 rads                          A  ≈( pi/2 - .365)rads ≈  1.205 rads

 

Because of squaring....some of these solutions are extraneous.....as this graph shows : https://www.desmos.com/calculator/ucfunqh12v

 

The correct solutions from [0, 2pi]   are

 

A  ≈ [ .365 , 1.206, 2.94, 4.91 ] rads

 

 

 

cool cool cool

 Sep 28, 2019

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