Respuesta :

dalendrk
[tex]2\sin^2x-\sin x=1\ \ \ \ |subtract\ 1\ from\ both\ sides\\\\2\sin^2x-\sin x-1=0\ \  \ \ |subtitute\ \sin x=t\ where\ t\in[-1;\ 1]\\\\2t^2-t-1=0\\\\2t^2-2t+t-1=0\\\\2t(t-1)+1(t-1)=0\\\\(t-1)(2t+1)=0\iff t-1=0\ or\ 2t+1=0\\\\t=1\ or\ 2t=-1\to t=-\dfrac{1}{2}\\\\therefore\\\sin x=1\ or\ \sin x=-\dfrac{1}{2}\\\\Answer:\boxed{t\in\left\{\dfrac{\pi}{2};\ \dfrac{7\pi}{6};\ \dfrac{11\pi}{6}\right\}}[/tex]

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