Question

Solve the following inequality:
$\sin x(\cos x+\frac{1}{2})⩽0$

Answer 1

$\sin x(\cos x+\frac{1}{2})\le 0$

Case i) $\sin x\le 0$ and $\cos \ge \frac{-1}{2}$

for $\sin x\le 0$, we get

$xϵ((2n-1)\pi ,2n\pi )$

for $\cos x\ge \frac{-1}{2}$, we get

$xϵ(2n\pi -\frac{2\pi }{3},2n\pi +\frac{2\pi }{3})$

Now combining, we get

$xϵ((2n\pi -\frac{2\pi }{3}),2n\pi )$

$xϵ[2n\pi -\frac{2\pi }{3},2n\pi ]$

Case ii)

$\sin x\ge 0$ and $\cos x\le \frac{-1}{2}$

for $\sin x\ge 0$, we get

$xϵ(\left(2n\right)\pi ,(2n+1)\pi )$

for $\cos x\le \frac{-1}{2}$, we get

$xϵ((2n+1)\pi -\frac{\pi }{3},(2n+1)\pi +\frac{\pi }{3})$

Combining, we get

$xϵ[(2n+1)\pi -\frac{\pi }{3},(2n+1)\pi ]$

Now for total cases,

$xϵ[(2n\pi -\frac{2\pi }{3}),2n\pi ]\cup [(2n+1)\pi -\frac{\pi }{3},(2n+1)\pi ]$