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Which is a solution to the equation (√3tanθ-3)(8cosx+8)=0 where 0≤θ≤2π? NOTE** It's root3 tan, the tan is not under the radical

 

Options: 

π/2

11π/6 

π/3 

 

I chose π/3

 Dec 13, 2017
edited by Julius  Dec 13, 2017
 #1
avatar+99276 
+1

(sqrt(3)  tan x   - 3)  ( 8cos x  + 8)  = 0

 

Either

 

sqrt (3)  tan x  -  3    = 0     add 3 to both sides

 

sqrt (3)   tan x   =  3        divide both sides by sqrt (3)

 

tan x  =  sqrt (3)   and this happens at    pi/3 and at 4pi/3

 

Or

 

8 cos x  +  8  = 0     subtract 8 from both sides

 

8 cos x  = -8     divide both sides by 8

 

cos x  = -1    and this happens at  pi

 

 

pi/3  is correct   !!!!!

 

 

cool cool cool

 Dec 13, 2017

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