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Question

Mathematics Question on Trigonometric Functions

If cos x = |sin x| then, the general solution is

A

x=nπ+(1)nπ4,nZx = n \pi + (-1)^n \frac{\pi}{4} , n \in Z

B

x=nπ±π4,nZx = n \pi \pm \frac{\pi}{4} , n \in Z

C

x=(2n1)π±π4,nZx = (2n - 1) \pi \pm \frac{\pi}{4} , n \in Z

D

x=2nπ±π4,nZx = 2n \pi \pm \frac{\pi}{4} , n \in Z

Answer

x=2nπ±π4,nZx = 2n \pi \pm \frac{\pi}{4} , n \in Z

Explanation

Solution

cosx=sinxcos\, x=\left|sin\, x\right|
±cosx=sinx\Rightarrow\, \pm cos x=sin x
tanx=±1\Rightarrow\, tan x=\pm1
x=nπ±π4,nZ,butcosxispositivesox=2nπ±π4,nZ.x=n\pi \pm\frac{\pi}{4}, n\in Z, but \,cos\, x \,is\, positive\, so \,x\,=2n \pi\pm\frac{\pi}{4}, n\in Z.