The solution of the equation (sin x + cos x)^ | vedclass

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

Question

(C) Given the equation $(\sin x + \cos x)^{1 + \sin 2x} = 2$. Since $1 + \sin 2x = (\sin x + \cos x)^2$,the equation becomes $(\sin x + \cos x)^{(\sin

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Given the equation $(\sin x + \cos x)^{1 + \sin 2x} = 2$. Since $1 + \sin 2x = (\sin x + \cos x)^2$,the equation becomes $(\sin x + \cos x)^{(\sin x + \cos x)^2} = 2$. Let $u = \sin x + \cos x$. Then $u^{u^2} = 2$. We know that $-\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2}$,so $-\sqrt{2} \leq u \leq \sqrt{2}$. If $u = \sqrt{2}$,then $(\sqrt{2})^{(\sqrt{2})^2} = (\sqrt{2})^2 = 2$. This is a solution. For $u = \sqrt{2}$,$\sin x + \cos x = \sqrt{2} \implies \frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = 1 \implies \sin(x + \frac{\pi}{4}) = 1$. Thus,$x + \frac{\pi}{4} = \frac{\pi}{2} \implies x = \frac{\pi}{4}$. If $u = -\sqrt{2}$,then $(-\sqrt{2})^{(-\sqrt{2})^2} = (-\sqrt{2})^2 = 2$. This is also a solution. For $u = -\sqrt{2}$,$\sin x + \cos x = -\sqrt{2} \implies \sin(x + \frac{\pi}{4}) = -1$. Thus,$x + \frac{\pi}{4} = -\frac{\pi}{2} \implies x = -\frac{3\pi}{4}$. Comparing with the given options,$x = \frac{\pi}{4}$ is the correct choice.

The solution of the equation $(\sin x + \cos x)^{1 + \sin 2x} = 2$,where $-\pi \leq x \leq \pi$,is

Similar Questions

The number of solutions in $[0, 2\pi]$ of the equation $16^{\sin^2 x} + 16^{\cos^2 x} = 10$ is

If $6 \cos 2 \theta + 2 \cos^2 \left(\frac{\theta}{2}\right) + 2 \sin^2 \theta = 0$ for $-\pi < \theta < \pi$,then $\theta =$

If $\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = 3$,then the general value of $\theta$ is

The number of values of $x$ with $0 \leq x \leq 2 \pi$ satisfying the equation $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$ is

The general solution of $|\sin x|=\cos x$ is (when $n \in I$) given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module