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

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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

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