The solution set of inequality $\cos x \geq-\frac{1}{2}$ is

$\bigcup_\limits {n \in Z}\left(n \pi-\frac{2 \pi}{3}, n \pi+\frac{2 \pi}{3}\right)$

$\bigcup_\limits {n \in Z}\left(2 n \pi-\frac{2 \pi}{3}, n \pi-\frac{\pi}{3}\right)$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Hard

1 of 1

Understanding +4 | -1

The solution set of inequality

\cos x \geq-\frac{1}{2}

View answer and explanation

Correct option: A

\bigcup\limits_{{n \in Z}}\left[2 n \pi-\frac{2 \pi}{3}, 2 n \pi+\frac{2 \pi}{3}\right]

Option A (

\bigcup\limits_ {{n \in Z}}\left[2 n \pi-\frac{{2 \pi}}{3}, 2 n \pi+\frac{{2 \pi}}{3}\right]

) is correct because

\cos x \geq -1/2

when

x

is between

-2\pi/3

and

2\pi/3

in each period.

Practice All Chapter MCQs →

\sin \,\,x\,\,\;cos\,\,3x = \sin \,\,3x\,\,\cos \,\,5x

\left[ {0,\;\,\frac{\pi }{2}} \right]

\left( {x,\,y} \right)

x,\,\,y \in \left[ {0,10} \right]

\left( {\sqrt {{{\sin }^2} - \sin \,\,x + \frac{1}{2}} } \right){2^{{{\sec }^2}y}}\,\, \cdot \,\, \le 1

\mathrm{X}

\theta \rightarrow \frac{\pi}{2}

\sin x+\cos x=\frac{\sqrt{7}}{2}

x \in\left[0, \frac{\pi}{4}\right]

\tan \frac{x}{2}

\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}

0 < x < \pi

\cos x+\sin x=\frac{1}{2}

\tan x