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Question Number 147581 by EDWIN88 last updated on 22/Jul/21

$$2\sin 2\mathrm{x}-4\sin {}^2\mathrm{x}=7\cos 2\mathrm{x}$$$$\frac{\pi }{2}<\mathrm{x}<\pi \Rightarrow \sin 2\mathrm{x}=?$$

Answered by iloveisrael last updated on 22/Jul/21

$$\mathrm{If}2\sin 2\mathrm{x}-4\sin {}^2\mathrm{x}=7\cos 2\mathrm{x}$$$$\mathrm{where}\frac{\pi }{2}<\mathrm{x}<\pi$$$$\mathrm{Then}\sin 2\mathrm{x}=?$$$$\pi <\mathrm{x}<2\pi \Rightarrow \sin 2\mathrm{x}<0$$$$2\sin 2\mathrm{x}-4\left(\frac{1-\cos 2\mathrm{x}}{2}\right)=7\cos 2\mathrm{x}$$$$2\sin 2\mathrm{x}-2+2\cos 2\mathrm{x}=7\cos 2\mathrm{x}$$$$2\sin 2\mathrm{x}-5\cos 2\mathrm{x}-2=0$$$$2\tan 2\mathrm{x}-5=2\sec 2\mathrm{x}$$$$4\tan {}^{2}2\mathrm{x}-20\tan 2\mathrm{x}+25=4+4\tan {}^{2}2\mathrm{x}$$$$20\tan 2\mathrm{x}=21$$$$\tan 2\mathrm{x}=\frac{21}{20}\Rightarrow \sin 2\mathrm{x}=-\frac{21}{\sqrt{{21}^2+{20}^2}}$$$$\sin 2\mathrm{x}=-\frac{20}{\sqrt{841}}=-\frac{20}{29}.$$

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