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Question Number 18524 by Tinkutara last updated on 23/Jul/17

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{equation}$$$${\sin }^{3}x-3\sin x{\cos }^{2}x+{2\cos }^{3}x=0\mathrm{in}$$$$[-\frac{\pi }{4},\frac{\pi }{4}]\mathrm{is}$$

Answered by Tinkutara last updated on 28/Jul/17

$$\mathrm{Using}\mathrm{AM}⩾\mathrm{GM}$$$$\frac{{\sin }^{3}x+{\cos }^{3}x+{\cos }^{3}x}{3}⩾\sin x{\cos }^{2}x$$$${\sin }^{3}x+{\cos }^{3}x+{\cos }^{3}x⩾3\sin x{\cos }^{2}x$$$$\therefore \sin x=\cos x\Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4}$$$$\mathrm{Hence}1\mathrm{solution}\mathrm{in}[-\frac{\pi }{4},\frac{\pi }{4}].$$

Answered by ajfour last updated on 29/Jul/17

$$\sin {}^3\mathrm{x}-\sin \mathrm{xcos}{}^2\mathrm{x}-2\sin \mathrm{xcos}{}^2\mathrm{x}$$$$+2\cos {}^3\mathrm{x}=0$$$$\sin \mathrm{x}(\sin {}^2\mathrm{x}-\cos {}^2\mathrm{x})-2\cos {}^2\mathrm{x}(\sin \mathrm{x}-\cos \mathrm{x})=0$$$$(\sin \mathrm{x}-\cos \mathrm{x})(\sin {}^2\mathrm{x}+\sin \mathrm{xcos}\mathrm{x}-2\cos {}^2\mathrm{x})=0$$$$(\sin \mathrm{x}-\cos \mathrm{x})(\sin {}^2\mathrm{x}-\sin \mathrm{xcos}\mathrm{x}$$$$+2\sin \mathrm{xcos}\mathrm{x}-2\cos {}^2\mathrm{x})=0$$$$\Rightarrow (\sin \mathrm{x}-\cos \mathrm{x})(\sin \mathrm{x}-\cos \mathrm{x})$$$$\times (\sin \mathrm{x}-2\cos \mathrm{x})=0$$$${(\sin \mathrm{x}-\cos \mathrm{x})}^2(\sin \mathrm{x}-2\cos \mathrm{x})=0$$$$\Rightarrow \sin \mathrm{x}=\cos \mathrm{x}\mathrm{or}$$$$\sin \mathrm{x}=2\cos \mathrm{x}$$$$\mathbf{I}\mathrm{n}[-\frac{\pi }{4},\frac{\pi }{4}]\mathrm{only}\mathrm{solution}\mathrm{is}\mathrm{x}=\frac{\pi }{4}.$$

Commented by Tinkutara last updated on 29/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!\mathrm{Also}\mathrm{a}\mathrm{nice}\mathrm{method}.$$

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