The general solution of sin x + cos x = 1 is | vedclass

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

Question

(C) Given equation: $\sin x + \cos x = 1$ Divide both sides by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given equation: $\sin x + \cos x = 1$ Divide both sides by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$ $\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$: $\sin(x + \frac{\pi}{4}) = \sin \frac{\pi}{4}$ The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$. Therefore,$x + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$ $x = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}, n \in Z$.

The general solution of $\sin x + \cos x = 1$ is

Similar Questions

The number of solutions of the equation $2 \theta - \cos^{2} \theta + \sqrt{2} = 0$ is equal to

The set $\{x \in R: \cos 2x + 2 \cos^2 x = 2\}$ is equal to

Find the general solution of $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$.

The smallest positive angle which satisfies the equation $2\sin^2 \theta + \sqrt{3} \cos \theta + 1 = 0$ is

If $n$ is any integer,then the general solution of the equation $\cos x - \sin x = \frac{1}{\sqrt{2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module