Question

Q. If$\theta \in \left(0,2\mathrm{\pi }\right)and2\sin \theta -5\sin \theta +2>0thenrangeof\theta is\left(1\right)\left(0,\mathrm{\pi }/6\right)\cup (5\pi /6,2\pi )\left(2\right)(0,\pi /6)\cup (\mathrm{\pi },2\mathrm{\pi })\left(3\right)(0,\mathrm{\pi }/6)\cap (\mathrm{\pi },2\mathrm{\pi })\left(4\right)0$

Answer 1

Dear student

$$\begin{gathered} 2{\sin }^2\mathrm{\theta }-5\mathrm{sin\theta }+2>0 \\ \Rightarrow \left(\mathrm{sin\theta }-2\right)\left(2\mathrm{sin\theta }-1\right)>0 \\ \Rightarrow \mathrm{sin\theta }-2>0\mathrm{or}2\mathrm{sin\theta }-1>0 \\ \mathrm{sin\theta }<\frac{1}{2}\mathrm{and}\mathrm{sin\theta }-2>0(\mathrm{Not}\mathrm{possible}) \\ \mathrm{So},\mathrm{\theta }\in \left(0,\frac{\mathrm{\pi }}{6}\right)\cup \left(\frac{5\mathrm{\pi }}{6},2\mathrm{\pi }\right) \end{gathered}$$
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