Question

The complete set of values of $x$ satisfying $\sin x>\frac{1}{2}$ is

• A. $2n\pi <x<2n\pi +\frac{\pi }{6},n\in Z$
• B. $2n\pi +\frac{\pi }{6}<x<2n\pi +\frac{5\pi }{6},n\in Z$
• C. $2n\pi +\frac{\pi }{6}<x<2n\pi +\frac{\pi }{2},n\in Z$
• D. None of these

Answer 1

The correct option is B $2n\pi +\frac{\pi }{6}<x<2n\pi +\frac{5\pi }{6},n\in Z$

When $\sin x=\frac{1}{2},$ we have
$x=\frac{\pi }{6},\frac{5\pi }{6}$ for $x\in [0,2\pi ]$


From the graph, it is clear that $\sin x>\frac{1}{2}$ when $x\in (\frac{\pi }{6},\frac{5\pi }{6})$
$\Rightarrow 2n\pi +\frac{\pi }{6}<x<2n\pi +\frac{5\pi }{6},n\in Z$