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Question Number 158354 by cortano last updated on 03/Nov/21

$$\sqrt{\sin (\frac{\pi }{4}+x)}+\sqrt{\sin (\frac{\pi }{4}-x)}=\sqrt[4]{2\cos 2x}$$

Commented by tounghoungko last updated on 03/Nov/21

$$\sin (\frac{\pi }{4}+x)+\sin (\frac{\pi }{4}-x)+2\sqrt{\sin (\frac{\pi }{4}+x)\sin (\frac{\pi }{4}-x)}=\sqrt{2\cos 2x}$$$$\left(i\right)\sin (\frac{\pi }{4}+x)+\sin (\frac{\pi }{4}-x)=2\sin \frac{\pi }{4}\cos x=\sqrt{2}\cos x$$$$\left(ii\right)\sin (\frac{\pi }{4}+x)\sin (\frac{\pi }{4}-x)=-\frac{1}{2}(\cos \frac{\pi }{2}-\cos 2x)=\frac{\cos 2x}{2}$$$$\Rightarrow \sqrt{2}\cos x+2\sqrt{\frac{\cos 2x}{2}}=\sqrt{2\cos 2x}$$$$\Rightarrow \sqrt{2}\cos x+\sqrt{2\cos 2x}=\sqrt{2\cos 2x}$$$$\Rightarrow \cos x=0\Rightarrow x=\frac{\pi }{2}$$$$checkRHS\Rightarrow \sqrt[4]{2\cos 2x}=\sqrt[4]{2\cos \pi }=\sqrt[4]{-2}\notin \mathbb{R}$$$$sotheequationhasnosolution$$

Commented by MJS_new last updated on 03/Nov/21

$$\mathrm{with}x=n\pi +\frac{\pi }{2}\mathrm{lhs}=\mathrm{rhs}=\frac{1}{\sqrt[4]{2}}(1+\mathrm{i})\Rightarrow \mathrm{you}$$$$\mathrm{found}\mathrm{the}\mathrm{solution}!$$

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