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Question Number 158774 by HongKing last updated on 08/Nov/21

Answered by MJS_new last updated on 08/Nov/21

$${\sin }^{4}x+{\cos }^{2}x={\sin }^{2}x+{\cos }^{4}x$$$$\Rightarrow \mathrm{we}\mathrm{have}$$$$2\sqrt{7+\cos 4x}+\sqrt{9-\cos 4x}=6\sqrt{2}$$$$\mathrm{squaring}\mathrm{\&}\mathrm{transforming}2\mathrm{times}$$$${\cos }^{2}4x-\frac{242}{25}\cos 4x+\frac{217}{25}=0$$$$\cos 4x=\frac{217}{25}\left[\mathrm{impossible}\right]$$$$\cos 4x=1\Rightarrow x=\frac{n\pi }{2}\wedge n\in \mathbb{Z}$$

Commented by HongKing last updated on 09/Nov/21

$$\mathrm{perfect}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathbf{S}\mathrm{er}$$

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