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Question

Mathematics Question on Trigonometric Functions

\left\\{x\,\in\,R : \left| \cos\,x\right|\ge \sin\,x\right\\} \cap\left[0, \frac{3\pi}{2}\right]=

A

[0,π4][3π4,3π2]\left[0, \frac{\pi}{4}\right]\cup\left[\frac{3\pi}{4}, \frac{3\pi}{2}\right]

B

[0,π4][π2,3π2]\left[0, \frac{\pi}{4}\right]\cup\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]

C

[0,π4][5π4,3π2]\left[0, \frac{\pi}{4}\right]\cup\left[\frac{5\pi}{4}, \frac{3\pi}{2}\right]

D

[0,3π2]\left[0, \frac{3\pi}{2}\right]

Answer

[0,π4][3π4,3π2]\left[0, \frac{\pi}{4}\right]\cup\left[\frac{3\pi}{4}, \frac{3\pi}{2}\right]

Explanation

Solution

Given, xR:cosxsinx[0,3π2]\\{x \in R:|\cos x| \geq \sin x\\} \cap\left[0, \frac{3 \pi}{2}\right]
If we draw the graphs of cosx|\cos x| and sinx\sin x, clearly
cosxsinx|\cos x| \geq \sin x when


x[0,π4][3π4,3π2]x \in\left[0, \frac{\pi}{4}\right] \cup\left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]
x[0,π4][3π4,3π2][0,3π2]\therefore x \in\left[0, \frac{\pi}{4}\right] \cup\left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right] \cap\left[0, \frac{3 \pi}{2}\right]
x[0,π4][3π4,3π2]\Rightarrow x \in\left[0, \frac{\pi}{4}\right] \cup\left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]