Question

Let $\alpha =\frac{\pi }{3},$ then the solution set of the inequality ${\log }_{\sin \alpha }(2-{\cos }^{2}x)<{\log }_{\sin \alpha }(1-\sin x),$ where $x\epsilon \left(0.2\pi \right)$ and $x\ne \frac{\pi }{2},$ is :

• A. $(\alpha ,\pi -\alpha )$
• B. $(\alpha ,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi -\alpha )$
• C. $(\frac{\pi }{3},\frac{5\pi }{6})$
• D. $(0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi )$

Answer 1

Answer: (D)

$(0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi )$

$\sin \alpha =\frac{1}{2}$

${\log }_{\sin \alpha }(2-{\cos }^{2}x)-{\log }_{\sin \alpha }(1-\sin x)<0$

${\log }_{\sin \alpha }\left\{\frac{2-{\cos }^{2}x}{1-\sin x}\right\}<0$

Since $\sin \alpha <1$

$\Rightarrow \frac{2-{\cos }^{2}x}{1-\sin x}>1$

$\Rightarrow \frac{1+1-{\cos }^{2}x}{1-\sin x}-1>0$

$\Rightarrow \frac{1+{\sin }^{2}x}{1-\sin x}-1>0\Rightarrow \frac{{\sin }^{2}x+\sin x}{1-\sin x}>0$

$\Rightarrow \frac{\sin x(\sin x+1)}{\sin x-1}<0$

$\Rightarrow \frac{\sin x}{\sin x-1}<0$ ($\because \left(\sin x\right)+1$ is always positive)

$\Rightarrow \sin xϵ(0,1)$

$\Rightarrow xϵ(0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi )$