`~rai~`
\(\cos^2x+\sin x-1=0\\\Leftrightarrow \cos^2x+\sin x-\sin^2x-\cos^2x=0\\\Leftrightarrow \sin x-\sin^2x=0\\\Leftrightarrow \sin x(1-\sin x)=0\\\Leftrightarrow \left[\begin{array}{I}\sin x=0\\\sin x=1\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x_1=k\pi\\x_2=\dfrac{\pi}{2}+k2\pi\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Do }x\in(0;\pi)\\TH1:0<x_1<\pi\\\Leftrightarrow 0<k\pi<\pi\\\Leftrightarrow 0<k<1\\\text{mà }k\in\mathbb{Z}\Rightarrow k\in\varnothing\Rightarrow x\in\varnothing.\\TH2:0<x_2<\pi\\\Leftrightarrow 0<\dfrac{\pi}{2}+k2\pi<\pi\\\Leftrightarrow -\dfrac{\pi}{2}<k2\pi<\dfrac{\pi}{2}\\\Leftrightarrow -\dfrac{\pi}{4}<k<\dfrac{\pi}{4}\\\text{mà }k\in\mathbb{Z}\Rightarrow k=0\Rightarrow x=\dfrac{\pi}{2}.\\\text{Vậy phương trình có nghiệm duy nhất trên khoảng }(0;\pi)\text{ là }x=\dfrac{\pi}{2}.\)
