Question

GPT: \(2\sin\left(2x+\dfrac{\pi}{4}\right)=\sqrt{2}\)

Answer 1

\(\Leftrightarrow\sin\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k2\pi\\2x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\left(k\in Z\right)\)

Answer 2

`2sin(2x+\pi/4)=\sqrt{2}`

`<=>sin(2x+\pi/4)=\sqrt{2}/2`

`<=>` $\left[\begin{matrix} 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\ 2x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}\end{matrix}\right.$    `(k in ZZ)`

`<=>` $\left[\begin{matrix} 2x=k2\pi\\ 2x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.$  `(k in ZZ)`

`<=>` $\left[\begin{matrix} x=k\pi\\ x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.$   `(k in ZZ)`