Solution of the equation [tex]\sqrt{3} (cosec\theta) -2=0[/tex]in the [ 0, 2π) is [tex]\frac{\pi }{3}[/tex] and [tex]\frac{2\pi }{3}[/tex].
" Trigonometric ratios are defined as relation of the ratio of the sides of the triangle to the acute angle of the given triangle enclosed in it."
Formula used
[tex]cosec\theta = \frac{1}{sin\theta}[/tex]
According to the question,
Given trigonometric ratio equation,
[tex]\sqrt{3} (cosec\theta) -2=0[/tex]
Replace trigonometric ratio [tex]cosec\theta[/tex] by [tex]sin\theta[/tex] in the above equation we get,
[tex]\sqrt{3} (\frac{1}{sin\theta} ) -2=0\\\\\implies \sqrt{3} (\frac{1}{sin\theta} ) = 2\\\\\implies sin\theta=\frac{\sqrt{3} }{2}[/tex]
As per given condition of the interval [ 0, 2π) we have,
[tex]\theta = sin^{-1} \frac{\sqrt{3} }{2} \\\\\ implies \theta = \frac{\pi }{3} or \frac{2\pi }{3}[/tex]
Hence, solution of the equation [tex]\sqrt{3} (cosec\theta) -2=0[/tex]in the [ 0, 2π) is
[tex]\frac{\pi }{3}[/tex] and [tex]\frac{2\pi }{3}[/tex].
Learn more about trigonometric ratio here
https://brainly.com/question/1201366
#SPJ3
