Question

find all solutions to the equation.

\\( \csc ^ { 2 } x - 2 = 0 \\)

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

example: \\( x = \frac { \pi } { 5 } + 2 k \pi , k \in \mathbb { z } \\) or \\( x = \frac { \pi } { 7 } + k \pi , k \in \mathbb { z } \\)

Explanation:

Step1: Solve for \(\csc x\)

Given \(\csc^{2}x - 2=0\), add \(2\) to both sides: \(\csc^{2}x=2\). Then take the square - root of both sides: \(\csc x=\pm\sqrt{2}\).
Since \(\csc x=\frac{1}{\sin x}\), we have \(\frac{1}{\sin x}=\pm\sqrt{2}\), so \(\sin x = \pm\frac{1}{\sqrt{2}}=\pm\frac{\sqrt{2}}{2}\).

Step2: Find the general solutions for \(\sin x=\frac{\sqrt{2}}{2}\) and \(\sin x =-\frac{\sqrt{2}}{2}\)

We know that \(\sin x=\frac{\sqrt{2}}{2}\) when \(x = \frac{\pi}{4}+ 2k\pi,k\in\mathbb{Z}\) or \(x=\frac{3\pi}{4}+2k\pi,k\in\mathbb{Z}\) (using the unit - circle, where \(\sin\theta=y\) coordinate of the point on the unit - circle \((\cos\theta,\sin\theta)\)).
Also, \(\sin x=-\frac{\sqrt{2}}{2}\) when \(x=\frac{5\pi}{4}+2k\pi,k\in\mathbb{Z}\) or \(x=\frac{7\pi}{4}+2k\pi,k\in\mathbb{Z}\).

We can combine these solutions. The general solution for \(\sin x=\pm\frac{\sqrt{2}}{2}\) is \(x=\frac{\pi}{4}+k\pi,k\in\mathbb{Z}\)

Answer:

\(x=\frac{\pi}{4}+k\pi,k\in\mathbb{Z}\)