The solutions to the trigonometric equation in the desired interval are given as follows:
[tex]\theta = \frac{\pi}{3}, \theta = \frac{5\pi}{3}[/tex]
The trigonometric equation is given by:
[tex]\sqrt{3}\cot{\theta} - 1 = 0[/tex]
Solving it similarly to an equation, we have that:
[tex]\sqrt{3}\cot{\theta} = 1[/tex]
[tex]\cot{\theta} = \frac{1}{\sqrt{3}}[/tex]
Since [tex]\cot{\theta} = \frac{1}{\tan{\theta}}[/tex], we have that the equation is equivalent to:
[tex]\tan{\theta} = \sqrt{3}[/tex]
The tangent is positive in the first and in the fourth quadrant. In the first quadrant, the angle [tex]\theta[/tex] with [tex]\tan{\theta} = \sqrt{3}[/tex] is:
[tex]\theta = \frac{\pi}{3}[/tex]
In the fourth quadrant, the equivalent angle is:
[tex]\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}[/tex]
More can be learned about trigonometric equations at https://brainly.com/question/24680641
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