SOLUTION: What value of x in the interval 0° ≤ x ≤ 180° satisfies the equation sqrt3 tan x + 1 = 0

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Question 443396: What value of x in the interval 0° ≤ x ≤ 180° satisfies the equation sqrt3 tan x + 1 = 0
Found 2 solutions by swincher4391, stanbon:
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%283%29+%2A+tan%28x%29+%2B1+=+0
sqrt%283%29+%2Atan%28x%29+=+-1
tan%28x%29+=+%28-1%2Fsqrt%283%29%29
Where is tangent negative? In quadrants 2 and 4.
But we are restricted to quadrants 1 and 2, since 0 < x < 180.
Then our angle must be in quadrant 2.
set up a 30,60,90 triangle.
The side opposite angle 30 is 1.
The side opposite angle 60 is sqrt(3)
The hypotenuse is 2.
So tangent is defined as opp/adj. So find where our opposite is 1. That's at angle 30.
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Now remember we are in quadrant 2.
So we need to create a reference angle.
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Reference angles are only based off of those that lie on the x-axis.
That's 0, 180, 360....etc.
Since 90 < x <180, 180 is what we will base our reference angle from.
Take 180 and subtract 30 (our reference angle) to get 150.
highlight%28150%29 is our answer.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
What value of x in the interval 0° ≤ x ≤ 180°
satisfies the equation sqrt3 tan x + 1 = 0
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sqrt(3)tan(x) = -1
tan(x) = -1/sqrt(3)
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Take the tan^-1 of both sides to get:
x = -30 degrees
That angle is in the 4th Quadrant.
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For your problem the equivalent angle in the 2nd Quadrant
would be x = 180-30 = 150 degrees
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Cheers,
Stan H.
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