Question

Solve: $2{\sin }^2\theta -5\sin \theta +2>0,ϵ[0,2\pi ]$

Answer 1

$2{\sin }^2\theta -5\sin \theta +2>0$ $0ϵ[0,2\pi ]$

$2{\sin }^2\theta -4\sin \theta -\sin \theta +2>0$

$2\sin \theta (\sin \theta -2)-1(\sin \theta -2)>0$

$(\sin \theta -2)(2\sin \theta -1)>0$

$(\sin \theta -2)$ is always negative

$(2\sin \theta -1)<0$

$\sin \theta <\frac{1}{2}$

$ln[0,2\pi ]$

$0ϵ[0,\frac{\pi }{6})\cup (5\frac{\pi }{6},2\pi ]$