The general solution of cos theta - sin theta | vedclass

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

Question

(A) Divide the equation $\cos 	heta - \sin 	heta = 1$ by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \cos 	heta - \frac{1}{\sqrt{2}} \sin 	heta = \frac{1}{\sqr

Vedclass · Accepted Answer

$	heta = 2n\pi - \frac{\pi}{2}, 	heta = 2n\pi, n \in Z$

Vedclass · Answer

(A) Divide the equation $\cos 	heta - \sin 	heta = 1$ by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \cos 	heta - \frac{1}{\sqrt{2}} \sin 	heta = \frac{1}{\sqrt{2}}$$\cos 	heta \cos \frac{\pi}{4} - \sin 	heta \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$$\cos(	heta + \frac{\pi}{4}) = \cos \frac{\pi}{4}$The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$:$	heta + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}$Case $1$: $	heta + \frac{\pi}{4} = 2n\pi + \frac{\pi}{4} \implies 	heta = 2n\pi$Case $2$: $	heta + \frac{\pi}{4} = 2n\pi - \frac{\pi}{4} \implies 	heta = 2n\pi - \frac{\pi}{2}$Thus, $	heta = 2n\pi, 2n\pi - \frac{\pi}{2}$ where $n \in Z$.

The general solution of $\cos \theta - \sin \theta = 1$ is

Similar Questions

The value of $\theta$ between $0^\circ$ and $360^\circ$ satisfying the equation $\tan \theta + \frac{1}{\sqrt{3}} = 0$ is equal to:

The number of solutions of the equation $\sin(9x) + \sin(3x) = 0$ in the closed interval $[0, 2\pi]$ is

The number of solutions of the equation $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$ in the interval $(0, \pi)$ is:

The solutions of the equation $\sin 2x + \cos 2x = 0$,where $\pi < x < 2\pi$,are

The number of values of $x$ in the interval $(0, 5 \pi)$ satisfying the equation $3 \sin^2 x - 7 \sin x + 2 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module