Question

solve the equation. give a general formula for all the solutions. list six solutions.

find all the solutions to ( sin (2 \theta)=-\frac{sqrt{2}}{2} ) in ( 0, pi) ).

( \theta=square ) (simplify your answer. type an exact answer, using ( pi ) as needed. use integers

Explanation:

Step1: Find the general solutions

We know that if \(\sin x =-\frac{\sqrt{2}}{2}\), then \(x = 2k\pi+\frac{5\pi}{4}\) or \(x = 2k\pi+\frac{7\pi}{4},k\in\mathbb{Z}\).
Since \(x = 2\theta\), we have \(2\theta=2k\pi+\frac{5\pi}{4}\) or \(2\theta=2k\pi+\frac{7\pi}{4}\).
Solving for \(\theta\), we get \(\theta=k\pi+\frac{5\pi}{8}\) or \(\theta=k\pi+\frac{7\pi}{8},k\in\mathbb{Z}\).

Step2: List six solutions

When \(k = 0\):
\(\theta=\frac{5\pi}{8}\) or \(\theta=\frac{7\pi}{8}\)
When \(k = 1\):
\(\theta=\pi+\frac{5\pi}{8}=\frac{13\pi}{8}\) or \(\theta=\pi+\frac{7\pi}{8}=\frac{15\pi}{8}\)
When \(k=- 1\):
\(\theta=-\pi+\frac{5\pi}{8}=-\frac{3\pi}{8}\) or \(\theta=-\pi+\frac{7\pi}{8}=-\frac{\pi}{8}\)

Answer:

The general solution is \(\theta = k\pi+\frac{5\pi}{8}\) or \(\theta=k\pi+\frac{7\pi}{8},k\in\mathbb{Z}\). Six solutions are \(-\frac{3\pi}{8},-\frac{\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8},\frac{13\pi}{8},\frac{15\pi}{8}\)