Đáp án:
$x=\dfrac{\pi^2}{9} $
Giải thích các bước giải:
$c)\sin \sqrt{x}=\dfrac{\sqrt{3}}{2}\left(-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}\right)\\ \text{ĐK: }x\ge 0 \Rightarrow x \in \left[0;\dfrac{\pi}{2}\right)\\ \Rightarrow \sqrt{x} \in \left[0;\sqrt{\dfrac{\pi}{2}}\right)\\ \sin \sqrt{x}=\dfrac{\sqrt{3}}{2}\\ \Leftrightarrow \sin \sqrt{x}=\sin \dfrac{\pi}{3}\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}=\dfrac{\pi}{3} + k 2 \pi(k \in \mathbb{Z})\\\sqrt{x}=\dfrac{2\pi}{3} + k 2 \pi(k \in \mathbb{Z})\end{array} \right.\\ \sqrt{x} \in \left[0;\sqrt{\dfrac{\pi}{2}}\right)\Leftrightarrow \left[\begin{array}{l} 0 \le \dfrac{\pi}{3} + k 2 \pi <\sqrt{\dfrac{\pi}{2}} (k \in \mathbb{Z})\\0 \le \dfrac{2\pi}{3} + k 2 \pi<\sqrt{\dfrac{\pi}{2}} (k \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} -\dfrac{\pi}{3} \le k 2 \pi <\sqrt{\dfrac{\pi}{2}}-\dfrac{\pi}{3} (k \in \mathbb{Z})\\\dfrac{-2\pi}{3} \le k 2 \pi<\sqrt{\dfrac{\pi}{2}}-\dfrac{2\pi}{3} (k \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow \left[\begin{array}{l} -\dfrac{1}{4} \le k <0,03 (k \in \mathbb{Z})\\-\dfrac{1}{3} \le k <-0,134(k \in \mathbb{Z})\end{array} \right.\\\Leftrightarrow k=0\\ \Rightarrow \sqrt{x}=\dfrac{\pi}{3} \\ \Rightarrow x=\dfrac{\pi^2}{9} $
