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

Values of $\theta$ $(0^\circ < \theta < 360^\circ)$ satisfying $\text{cosec } \theta + 2 = 0$ are

The solution set of the equation $\tan (\pi \tan x) = \cot (\pi \cot x)$ for $x \in (0, \frac{\pi}{2})$ is

If $n$ is the number of solutions of the equation $2 \cos x(4 \sin(\frac{\pi}{4}+x) \sin(\frac{\pi}{4}-x)-1)=1$ for $x \in [0, \pi]$,and $S$ is the sum of all these solutions,then the ordered pair $(n, S)$ is:

The number of solutions of the given equation $a \sin x + b \cos x = c$,where $|c| > \sqrt{a^2 + b^2}$,is

Find the general solution of the equation $\cos 4x = \cos 2x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module