Question

find all solutions to the equation.

√3 cscθ + 2 = 0

write your answer in radians in terms of π, and use the \or\ button as necessary.

example: θ = π/5 + 2kπ, k ∈ z or θ = π/7 + kπ, k ∈ z

θ = □

Explanation:

Step1: Isolate \(\csc\theta\)

Given \(\sqrt{3}\csc\theta + 2=0\), subtract \(2\) from both sides: \(\sqrt{3}\csc\theta=- 2\). Then divide by \(\sqrt{3}\): \(\csc\theta=-\frac{2}{\sqrt{3}}\). Since \(\csc\theta=\frac{1}{\sin\theta}\), we have \(\sin\theta=-\frac{\sqrt{3}}{2}\).

Step2: Find the general solutions

We know that \(\sin\theta =-\frac{\sqrt{3}}{2}\) when \(\theta=\frac{4\pi}{3}+2k\pi,k\in\mathbb{Z}\) (in the third - quadrant) or \(\theta=\frac{5\pi}{3}+2k\pi,k\in\mathbb{Z}\) (in the fourth - quadrant).

Answer:

\(\theta=\frac{4\pi}{3}+2k\pi,k\in\mathbb{Z}\text{ or }\theta=\frac{5\pi}{3}+2k\pi,k\in\mathbb{Z}\)