Provide the values of β that are the solutions to the equation:
2 cos ( 3 β + π 3 ) = − 2 2\cos\left(3\beta+\frac{\pi}{3}\right)=-\sqrt2 2 cos ( 3 β + 3 π ) = − 2
Solution: \textbf{Solution:} Solution:
2 cos ( 3 β + π 3 ) = − 2 2\cos\left(3\beta+\frac{\pi}{3}\right)=-\sqrt2 2 cos ( 3 β + 3 π ) = − 2
[ 3 β + π 3 = cos − 1 ( − 2 2 ) + 2 π k , k ∈ Z , 3 β + π 3 = − cos − 1 ( − 2 2 ) + 2 π k , k ∈ Z . \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. ⎣ ⎡ 3 β + 3 π = cos − 1 ( − 2 2 ) + 2 πk , k ∈ Z , 3 β + 3 π = − cos − 1 ( − 2 2 ) + 2 πk , k ∈ Z .
[ 3 β + π 3 = 3 π 4 + 2 π k , k ∈ Z , 3 β + π 3 = − 3 π 4 + 2 π k , k ∈ Z . \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. ⎣ ⎡ 3 β + 3 π = 4 3 π + 2 πk , k ∈ Z , 3 β + 3 π = − 4 3 π + 2 πk , k ∈ Z .
[ 3 β = 5 π 12 + 2 π k , k ∈ Z , 3 β = − 13 π 12 + 2 π k , k ∈ Z . \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. ⎣ ⎡ 3 β = 12 5 π + 2 πk , k ∈ Z , 3 β = − 12 13 π + 2 πk , k ∈ Z .
[ β = 5 π 36 + 2 π k 3 , k ∈ Z , β = − 13 π 36 + 2 π k 3 , k ∈ Z . \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. ⎣ ⎡ β = 36 5 π + 3 2 πk , k ∈ Z , β = − 36 13 π + 3 2 πk , k ∈ Z .
Answer: \textbf{Answer:} Answer:
β ∈ { 5 π 36 + 2 π k 3 } ∪ { − 13 π 36 + 2 π k 3 } , k ∈ Z \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}} β ∈ { 36 5 π + 3 2 πk } ∪ { − 36 13 π + 3 2 πk } , k ∈ Z
#340153 on Dec 2023