Question

Provide the values of β that are the solutions to the equation:

2cos(3β + π/3) = -√2

Answer 1

Provide the values of β that are the solutions to the equation:

$2\cos\left(3\beta+\frac{\pi}{3}\right)=-\sqrt2$.

$\textbf{Solution:}$

$2\cos\left(3\beta+\frac{\pi}{3}\right)=-\sqrt2$.

$\left[ \begin{gathered} 3\beta+\frac{\pi}{3}=\cos^{-1}\left(-\frac{\sqrt2}{2}\right)+2\pi k, \, k\in\mathbb {Z}, \\ 3\beta+\frac{\pi}{3}=-\cos^{-1}\left(-\frac{\sqrt2}{2}\right)+2\pi k, \, k\in\mathbb {Z}. \end{gathered} \right.$

$\left[ \begin{gathered} 3\beta+\frac{\pi}{3}=\frac{3\pi}{4}+2\pi k, \, k\in\mathbb {Z}, \\ 3\beta+\frac{\pi}{3}=-\frac{3\pi}{4}+2\pi k, \, k\in\mathbb {Z}. \end{gathered} \right.$

$\left[ \begin{gathered} 3\beta=\frac{5\pi}{12}+2\pi k, \, k\in\mathbb {Z}, \\ 3\beta=-\frac{13\pi}{12}+2\pi k, \, k\in\mathbb {Z}. \end{gathered} \right.$

$\left[ \begin{gathered} \beta=\frac{5\pi}{36}+\frac{2\pi k}{3}, \, k\in\mathbb {Z}, \\ \beta=-\frac{13\pi}{36}+\frac{2\pi k}{3}, \, k\in\mathbb {Z}. \end{gathered} \right.$

$\textbf{Answer:}$

$\boxed{\beta\in \{\frac{5\pi}{36}+\frac{2\pi k}{3} \}\cup \{-\frac{13\pi}{36}+\frac{2\pi k}{3}\}, \, k\in\mathbb {Z}}$