Hi Shades :)
2 sin2 x - 3 sin x + 1 = 0
\(2 sin^2 x - 3 sin x + 1 = 0\\ let\;\; y=sin^2x\\ 2y^2-3y+1=0\\ 2y^2-2y-y+1=0\\ 2y(y-1)-1(y-1)=0\\ (2y-1)(y-1)=0\\ 2y-1=0 \qquad or \qquad y-1=0\\ 2y=1 \qquad or \qquad y=1\\ y=\frac{1}{2} \qquad or \qquad y=1\\ so\\ sin^2x=\frac{1}{2} \qquad or \qquad sin^2x=1\\ sinx=\pm \frac{1}{\sqrt{2}} \qquad or \qquad sinx=\pm1\\\)
\(x=45^0,135^0,225^0,315^0, 90^0,\;\;or \;\;270^0\)
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