If n is any integer,then the general solution | vedclass

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

Question

(C) Given equation is $\cos x - \sin x = \frac{1}{\sqrt{2}}$.Dividing both sides by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\s

Vedclass · Accepted Answer

$x = 2n\pi + \frac{\pi}{12}$ or $x = 2n\pi - \frac{7\pi}{12}$

Vedclass · Answer

(C) Given equation is $\cos x - \sin x = \frac{1}{\sqrt{2}}$.Dividing both sides by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{2}$.This can be written as $\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4} = \frac{1}{2}$.Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get $\cos(x + \frac{\pi}{4}) = \frac{1}{2}$.Since $\cos \frac{\pi}{3} = \frac{1}{2}$,the equation becomes $\cos(x + \frac{\pi}{4}) = \cos \frac{\pi}{3}$.The general solution for $\cos 	heta = \cos \alpha$ is $	heta = 2n\pi \pm \alpha$.Therefore,$x + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{3}$.Case $1$: $x = 2n\pi + \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi + \frac{\pi}{12}$.Case $2$: $x = 2n\pi - \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi - \frac{7\pi}{12}$.

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

Similar Questions

The number of solutions to the equation $\cos^4 x + \frac{1}{\cos^2 x} = \sin^4 x + \frac{1}{\sin^2 x}$ in the interval $[0, 2\pi]$ is

The equation $\sqrt{3} \sin x + \cos x = 4$ has

Solve $\cos x = \frac{1}{2}$

Let $S$ be the sum of all solutions (in radians) of the equation $\sin^{4} \theta + \cos^{4} \theta - \sin \theta \cos \theta = 0$ in $[0, 4\pi]$. Then $\frac{8S}{\pi}$ is equal to ...... .

If $\sin^2 \theta - 2\cos \theta + \frac{1}{4} = 0$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module