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Question Number 209872 by Frix last updated on 24/Jul/24

$$\mathrm{Solve}\mathrm{for}x\in \mathbb{R}:$$$$x^3-3x^2+2=\sqrt{x+1}$$

Answered by Ghisom last updated on 24/Jul/24

$$x=t+1$$$$t^3-3t=\sqrt{t+2}$$$$\mathrm{expecting}3\mathrm{solutions}\mathrm{at}\mathrm{most}$$$$\mathrm{remember}\cos 3\alpha ={4\cos }^3\alpha -3\cos \alpha$$$$\Rightarrow t=2\cos \alpha \wedge 0⩽\alpha ⩽\pi$$$${8\cos }^3\alpha -6\cos \alpha =\sqrt{2+2\cos \alpha }$$$$2\cos 3\alpha =\sqrt{2(1+\cos \alpha )}$$$$\mathrm{remember}1+\cos 2\beta ={2\cos }^2\beta$$$$\Rightarrow$$$$2\cos 3\alpha =\sqrt{{4\cos }^2\frac{\alpha }{2}}$$$$\cos 3\alpha =\mid \cos \frac{\alpha }{2}\mid$$$$\Rightarrow \alpha =0\vee \alpha =\frac{4\pi }{7}\vee \alpha =\frac{4\pi }{5}$$$$\Rightarrow t=2\vee t=2\cos \frac{4\pi }{7}\vee t=-\frac{1+\sqrt{5}}{2}$$$$\Rightarrow x=3\vee x=1-2\cos \frac{4\pi }{7}\vee x=\frac{1-\sqrt{5}}{2}$$

Commented by Frix last updated on 24/Jul/24

$$\mathrm{Great}\mathrm{idea}!$$

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