Đáp án:
Giải thích các bước giải:
`\frac{1}{sin^2\ x}=2`
TXĐ: `D=\mathbb{R} \\ {k\pi\ (k \in \mathbb{Z})`
`⇔ 1=2sin^2 x`
`⇔ sin^2 x=1/2`
`⇔ sin\ x=\pm \frac{1}{\sqrt{2}}`
`⇔` \(\left[ \begin{array}{l}\sin\ x=\dfrac{\sqrt{2}}{2}\\\sin\ x=-\dfrac{\sqrt{2}}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{3\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{5\pi}{4}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\) (TM)
Vậy `S={\pm \frac{\pi}{4}+k2\pi\ (k \in \mathbb{Z});\frac{3\pi}{4}+k2\pi\ (k \in \mathbb{Z});\frac{5\pi}{4}+k2\pi\ (k \in \mathbb{Z})}`
