[tex]\it x\in(0,\ \pi) \Rightarrow sinx\in(0,\ 1) \Rightarrow sinx>0\ \ \ \ \ (1)\\ \\ sin(\pi-x)=sinx\ \ \ \ \ (2)\\ \\ 2sinx\ sin(\pi-x)=1\ \stackrel{(2)}{\Longrightarrow}\ 2sinx\cdot sinx=1 \Rightarrow sin^2x=\dfrac{1}{2} \Rightarrow \\ \\ \\ \Rightarrow \sqrt{sin^2x}=\sqrt{\dfrac{1}{2}}\ \stackrel{(1)}{\Longrightarrow}\ sinx=\sqrt{\dfrac{1}{2}}=\dfrac{^{\sqrt2)}1}{\sqrt2}=\dfrac{\sqrt2}{2} \Rightarrow\\ \\ \\ \Rightarrow x=\dfrac{\pi}{4},\ sau\ x=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}[/tex]
