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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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

Similar Questions

Let $S = \{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3} \sin\theta\}$. Then $\sum_{\theta \in S} \theta$ is equal to:

The number of solutions of the equation $\cos^2 2x + \cos^2 \frac{5x}{4} = \cos 2x \cos^2 5x$ in the interval $[0, \frac{\pi}{3}]$ is:

If $\sin \left(x+\frac{\pi}{3}\right)+\sin \left(x-\frac{\pi}{3}\right)=1$,then find the value of $x$ in the interval $[0, \pi]$.

If $n$ is the number of solutions of the equation $2 \cos x(4 \sin(\frac{\pi}{4}+x) \sin(\frac{\pi}{4}-x)-1)=1$ for $x \in [0, \pi]$,and $S$ is the sum of all these solutions,then the ordered pair $(n, S)$ is:

The number of solutions of the equation $\sec x = 1 + \cos x + \cos^2 x + \dots \infty$ in the interval $x \in [-50 \pi, 50 \pi]$ is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module