Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{4} + k2\pi\\x = \dfrac{3\pi}{4} + k2\pi\\x = -\dfrac{\pi}{4} + k2\pi\\x = \dfrac{5\pi}{4} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\left[\begin{array}{l}\sin^2x + \dfrac{1}{\sin^2x} = \dfrac{5}{2} \qquad (1)\\\sin^2x + \dfrac{1}{\sin^2x} = -\dfrac{7}{2}\quad (2)\end{array}\right.$
$ĐK:\, \sin x\ne 0 \Leftrightarrow x \ne k\pi$
Ta có:
$\sin^2x + \dfrac{1}{\sin^2x} > 0$
$\Rightarrow \sin^2x + \dfrac{1}{\sin^2x} > - \dfrac{7}{2}$
$\Rightarrow (2)$ vô nghiệm
$(1) \Leftrightarrow 2\sin^4x - 5\sin^2x + 2 = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin^2x = \dfrac{1}{2}\\\sin^2x = 2\quad (loại)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\sin x = \dfrac{\sqrt2}{2}\\\sin x = -\dfrac{\sqrt2}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{4} + k2\pi\\x = \dfrac{3\pi}{4} + k2\pi\\x = -\dfrac{\pi}{4} + k2\pi\\x = \dfrac{5\pi}{4} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
