Question

solve the inequality $cosx\le -\frac{1}{2}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Period of cos x is $2\pi$. That is why it is sufficient to solve inequality of the form $cosx\le -\frac{1}{2}$ first on the interval of $[0,2\pi ]$, and then get the solution set by adding numbers of the form $2n\pi$, $nϵz$ to the each solutions obtained in the interval.

Let us solve the inequality in the interval $[0,2\pi ]$

The graph of $y=cosxandy=-\left(\frac{1}{2}\right)$ are taken as two curves on x - y plane.

Darkened region is the part of $cosx\le -\frac{1}{2}$.

In interval $[0,2\pi ]$ $cosx=-\frac{1}{2}$ means

$x=\frac{2\pi }{3}and\frac{4\pi }{3}$

If $cosx\le -\frac{1}{2}$

$\frac{2\pi }{3}\le x\le \frac{4\pi }{3}$

On generating this solution ;

$2n\pi +\frac{2n}{3}\le x\le 2n\pi +\frac{4\pi }{3};nϵz$

$\therefore$ solution of $cosx\le -\frac{1}{2}$ is

$\Rightarrow xϵ[2n\pi +\frac{2\pi }{3},2n\pi +\frac{4\pi }{3}];nϵz$