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Question Number 18093 by Tinkutara last updated on 15/Jul/17

$$\mathrm{The}\mathrm{equation}\sin x+\sin 2x+2\sin x\sin 2x$$$$=2\cos x+\cos 2x\mathrm{is}\mathrm{satisfied}\mathrm{by}\mathrm{values}$$$$\mathrm{of}x\mathrm{for}\mathrm{which}$$$$\left(1\right)x=n\pi +{(-1)}^n\frac{\pi }{6},n\in I$$$$\left(2\right)x=2n\pi +\frac{2\pi }{3},n\in I$$$$\left(3\right)x=2n\pi -\frac{2\pi }{3},n\in I$$$$\left(4\right)x=2n\pi -\frac{\pi }{2},n\in I$$

Commented by ajfour last updated on 15/Jul/17

$$\mathrm{yes},\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{dear}.$$

Answered by ajfour last updated on 15/Jul/17

$$\sin \mathrm{x}+\sin 2\mathrm{x}+2\sin \mathrm{xsin}2\mathrm{x}$$$$=2\cos \mathrm{x}+\cos 2\mathrm{x}$$$$\sin \mathrm{x}+\sin 2\mathrm{x}+\cos \mathrm{x}-\cos 3\mathrm{x}$$$$=2\cos \mathrm{x}+\cos 2\mathrm{x}$$$$\Rightarrow \sin \mathrm{x}+2\sin \mathrm{xcos}\mathrm{x}$$$$=\cos \mathrm{x}+\cos 3\mathrm{x}+\cos 2\mathrm{x}$$$$\left(\sin \mathrm{x}\right)(1+2\cos \mathrm{x})=2\cos \mathrm{xcos}2\mathrm{x}$$$$+\cos 2\mathrm{x}$$$$\left(\sin \mathrm{x}\right)(1+2\cos \mathrm{x})-\left(\cos 2\mathrm{x}\right)(1+2\cos \mathrm{x})=0$$$$(1+2\cos \mathrm{x})(\sin \mathrm{x}-\cos 2\mathrm{x})=0$$$$(1+2\cos \mathrm{x})(\sin \mathrm{x}-1+2\sin {}^2\mathrm{x})=0$$$$(1+2\cos \mathrm{x})(1+\sin \mathrm{x})(2\sin \mathrm{x}-1)=0$$$$\Rightarrow \cos \mathrm{x}=-\frac{1}{2}\mathrm{and}/\mathrm{or}$$$$\sin \mathrm{x}=-1\mathrm{and}/\mathrm{or}\sin \mathrm{x}=\frac{1}{2}$$$$\Rightarrow \mathrm{x}=2\mathrm{n}\pi \pm \frac{2\pi }{3}$$$$\mathrm{x}=2\mathrm{n}\pi -\frac{\pi }{2}$$$$\mathrm{x}=\mathrm{n}\pi +{(-1)}^{\mathrm{n}}\frac{\pi }{6}.$$$$\mathrm{All}\mathrm{options}\mathrm{are}\mathrm{correct}.$$

Commented by Tinkutara last updated on 15/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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