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What are the solutions to the equation on [0,2π) ?

 

 

sec(3x)=−2√3/3

 Apr 9, 2020

Best Answer 

 #2
avatar+111329 
+1

sec (3x)  =  -2√3/3  =  -2/√3

 

Notice  that  this  is  the same as

 

cos (3x)  = -√3/2

 

Note  that  cos (x)  = -√3/2   at     (5/6)pi , (7/6)pi , (17/6)pi, (19/6)pi, (29/6)pi  and (31/6)pi

 

Dividing  each  of these angles  by 3  gives  us  the solutions we  want  =

 

(5/18)pi , (7/18)pi, (17/18)pi, (19/18) pi, (29/18)pi, (31/18)pi

 

 

cool cool cool

 Apr 9, 2020
 #1
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0

x = 5*pi/3 + 2*pi/3*k

 Apr 9, 2020
 #2
avatar+111329 
+1
Best Answer

sec (3x)  =  -2√3/3  =  -2/√3

 

Notice  that  this  is  the same as

 

cos (3x)  = -√3/2

 

Note  that  cos (x)  = -√3/2   at     (5/6)pi , (7/6)pi , (17/6)pi, (19/6)pi, (29/6)pi  and (31/6)pi

 

Dividing  each  of these angles  by 3  gives  us  the solutions we  want  =

 

(5/18)pi , (7/18)pi, (17/18)pi, (19/18) pi, (29/18)pi, (31/18)pi

 

 

cool cool cool

CPhill Apr 9, 2020

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