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Question Number 153079 by mathdanisur last updated on 04/Sep/21

$$\mathrm{Solve}\mathrm{the}\mathrm{equation}:$$$$\sqrt{1-\mathit{z}}=2{\mathit{z}}^2-1+2\mathit{z}\sqrt{1-{\mathit{z}}^2}$$

Commented by MJS_new last updated on 04/Sep/21

$$\mathrm{I}\mathrm{get}z=\sin \frac{\pi }{5}=\frac{\sqrt{10-2\sqrt{5}}}{4}$$

Answered by liberty last updated on 04/Sep/21

$$letz=\cos 2x$$$$\Rightarrow \sqrt{1-\cos 2x}=2\cos {}^{2}2x-1+2\cos 2x\sqrt{1-\cos {}^{2}2x}$$$$\Rightarrow \sqrt{2\sin {}^{2}x}=\cos 4x+2\cos 2x\sin 2x$$$$\Rightarrow \sqrt{2}\sin x=\cos 4x+\sin 4x$$$$\Rightarrow \sqrt{2}\sin x=\sqrt{2}\cos (4x-\frac{\pi }{4})$$$$\Rightarrow \cos (\frac{\pi }{2}-x)=\cos (4x-\frac{\pi }{4})$$$$nowiteasytofinish$$

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