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

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(A) Given $\cos x = |\sin x|$.Since $|\sin x| \ge 0$,we must have $\cos x \ge 0$,which implies $x$ lies in the first or fourth quadrant.Squaring both

Vedclass · Accepted Answer

$x = 2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z}$

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(A) Given $\cos x = |\sin x|$.Since $|\sin x| \ge 0$,we must have $\cos x \ge 0$,which implies $x$ lies in the first or fourth quadrant.Squaring both sides,we get $\cos^2 x = \sin^2 x$.This simplifies to $	an^2 x = 1$,so $	an x = \pm 1$.For $	an x = 1$,$x = n\pi + \frac{\pi}{4}$. For $	an x = -1$,$x = n\pi - \frac{\pi}{4}$.Combining these,$x = n\pi \pm \frac{\pi}{4}$.However,we must satisfy $\cos x \ge 0$. For $n$ even $(n=2k)$,$x = 2k\pi \pm \frac{\pi}{4}$,where $\cos x = \cos(\pm \frac{\pi}{4}) = \frac{1}{\sqrt{2}} > 0$ (Valid).For $n$ odd $(n=2k+1)$,$x = (2k+1)\pi \pm \frac{\pi}{4}$,where $\cos x = \cos(\pi \pm \frac{\pi}{4}) = -\frac{1}{\sqrt{2}} Thus,the general solution is $x = 2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z}$.

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

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