F asked in Science & MathematicsMathematics · 6 months ago

Find all solutions to the given equation in the interval [0,2π). Give the exact solution, including "pi" for π. 4cotx+4sqrt(3)=0?

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  • 6 months ago
    Favourite answer

    4cot(x)+4sqr(3)=0, 0=<x<2pi

    =>

    tan(x)=-1/sqr(3)

    =>

    x=5pi/6, or x=11pi/6

  • 6 months ago

    4cotx + 4√3 = 0

    so, cotx = -√3

    i.e. tanx = -1/√3....negative means Q2

    Then, x = -π/6 + nπ...for n = 0, 1, 2,...

    or, x = (π/6)(6n - 1)

    Hence, for n = 0 and 1 we have:

    x = -π/6, 5π/6 and 11π/6

    so, x = 5π/6 and 11π/6 are solutions for [0, 2π)

    :)> 

  • Vaman
    Lv 7
    6 months ago

    4cotx+4sqrt(3)=0,  cot x +sqrt 3=0, cot x +2 sqrt 3/2=0

    cot x + 2 sin 60=0, 1/2 cos x/ sin x +sin 60=0

    cos 60 sin x + sin 60 cos x=0

    sin(x+60)=0= sin pi. pi= x+60, x= 180-60=120. This is the answer.

  • 6 months ago

    Given --

    4 cot x + 4 √3  =  0

    Divide both the sides by 4 ---

    =>  cot x + √3  =  0

    => cot x  =  -  √3 

    => tan x  =  - ( 1/√3 )

    tan has negative values in 2nd and 4th Quadrants.

    =>  x  =  (5/6) pi  and (11/6) pi  ................ Answer

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  • ted s
    Lv 7
    6 months ago

    cot Θ = - √3 has the reference triangle for Θ of { - √ 3 ,  1 , 2 } for 2nd quadrant and { √3 , - 1 , 2 } for the 4th quadrant...5π / 6 and 11 π / 6

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