The general solution of sin x - cos x = sqrt{ | vedclass

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

Question

(A) Given,$\sin x - \cos x = \sqrt{2}$ Divide both sides by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = 1$ Using $\cos \frac{

Vedclass · Accepted Answer

$2n\pi + \frac{3\pi}{4}$

Vedclass · Answer

(A) Given,$\sin x - \cos x = \sqrt{2}$ Divide both sides by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x = 1$ Using $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get: $\sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} = 1$ Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $\sin(x - \frac{\pi}{4}) = 1$ Since $\sin 	heta = 1$ implies $	heta = 2n\pi + \frac{\pi}{2}$: $x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{2}$ $x = 2n\pi + \frac{\pi}{2} + \frac{\pi}{4}$ $x = 2n\pi + \frac{3\pi}{4}$

The general solution of $\sin x - \cos x = \sqrt{2}$,for any integer $n$ is

Similar Questions

If $2\sin^2 \theta = 3\cos \theta$,where $0 \le \theta \le 2\pi$,then $\theta = $

The general solution of $\cos x + \sin x = \cos 2x + \sin 2x$ is $x = np\pi$ or $x = \frac{nq\pi}{3} + \frac{\pi}{6}$ for $n \in \mathbb{Z}$. Then $p : q =$?

The possible values of $\theta \in (0, \pi)$ such that $\sin \theta + \sin (4 \theta) + \sin (7 \theta) = 0$ are

The number of solutions of $\tan x + \sec x = 2 \cos x$ in $[0, 2 \pi]$ are

If $\tan \theta - \sqrt{2} \sec \theta = \sqrt{3}$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module