2sin(2x+π/4)=1........ giúp mình với

Đáp án:

Giải thích các bước giải:

 `2sin\ (2x+\frac{\pi}{4})=1`

`⇔ sin\ (2x+\frac{\pi}{4})=1/2`

`⇔ sin\ (2x+\frac{\pi}{4})=sin\ \frac{\pi}{6}`

`⇔` \(\left[ \begin{array}{l}2x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\\2x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{6}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x=-\dfrac{\pi}{12}+k2\pi\ (k \in \mathbb{Z})\\2x=\dfrac{7\pi}{12}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=-\dfrac{\pi}{24}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{7\pi}{24}+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)