sa se rezolve ecuatiea trigonometrica:2sin la a 2x-sinx-1=0

[tex]\displaystyle\\ 2\sin^2x - \sin x-1=0\\\\ \text{Substitutie: } ~~~\boxed{y = \sin x}\\\\ 2y^2-y-1=0\\\\ y_{12}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}= \frac{1\pm \sqrt{1+8} }{4}= \frac{1\pm \sqrt{9} }{4}= \frac{1\pm 3 }{4}\\\\ y_1= \frac{1+ 3 }{4} = 1 \Longrightarrow \sin x=1\Longrightarrow \boxed{x_1 = \frac{\pi}{2}+2k\pi} \\\\ y_2=\frac{1-3}{4}\Rightarrow\sin x=-\frac{1}{2}\Rightarrow \boxed{x_2=\pi+\frac{\pi}{6}+2k\pi\bigcup2\pi-\frac{\pi}{6}+2k\pi} [/tex]