Question

Solution set of the inequality $\frac{2{\cos }^{2}x-1}{\sin \frac{x}{2}}<0$, can be.

• A. $\left(\frac{2n-1}{4}\right)\pi <x<\left(\frac{2n+1}{4}\right)\pi$, where $n=1,3,6$
• B. $\left(\frac{n}{4}\right)\pi <x<\left(\frac{n+1}{4}\right)\pi$, where $n=8$
• C. $\left(\frac{n-1}{4}\right)\pi <x<\left(\frac{n}{4}\right)\pi$, where $n=16$
• D. $\left(\frac{2n-1}{4}\right)\pi <x<\left(\frac{2n+1}{4}\right)\pi$, where $n=5$

Answer 1

The correct option is A $\left(\frac{2n-1}{4}\right)\pi <x<\left(\frac{2n+1}{4}\right)\pi$, where $n=1,3,6$