Question

Find the complete solution in radians of each equation. sinθcot²θ-3sin θ=0

Answer 1

Answer:
`\theta=(\pi)/(6)+\pi n, (5\pi)/(6)+\pi n, \pi n, n \in \mathbb{Z}`

Explanation:

`\sin \theta(\cot^2 \theta-3)=0\\\\\sin \theta(\cot \theta-\sqrt 3)(\cot \theta+\sqrt 3)=0\\\\\therefore \sin \theta=0, \cot \theta =\pm \sqrt 3\\\\\sin \theta =0 \implies \theta=\pi n, n \in \mathbb{Z}\\\\\cot \theta=\pm \sqrt 3 \implies \tan \theta =\pm (1)/(√(3)) \implies \theta=(\pi)/(6)+\pi n, (5\pi)/(6)+\pi n, n \in \double{Z}\\\\\therefore \theta=(\pi)/(6)+\pi n, (5\pi)/(6)+\pi n, \pi n, n \in \mathbb{Z}`