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Question Number 144829 by mathlove last updated on 29/Jun/21

$$\sin {}^{3}x+\cos {}^{3}x=1-\frac{1}{2}\sin {}^{2}x$$$$x=?$$

Answered by ajfour last updated on 29/Jun/21

$${(1-s^2)}^3={(1-\frac{s^2}{2}-s^3)}^2$$$$1-s^6+3s^4-3s^2$$$$=1+\frac{s^4}{4}+s^6-s^2+s^5-2s^3$$$$\Rightarrow s^6+\frac{s^5}{2}-\frac{11s^4}{8}-s^3+s^2=0$$$$\Rightarrow s=0or$$$$s^4+\frac{s^3}{2}-\frac{11s^2}{8}-s+1=0$$$$therearetworealpositive$$$$rootstothisquartic;say$$$$s=\alpha ,\beta$$$$now$$$$\sin x=0,\alpha ,\beta$$$$x=n\pi +{(-1)}^n\theta$$$$where\theta =0,\alpha ,\beta$$

Answered by liberty last updated on 29/Jun/21

$$\sin {}^3\mathrm{x}+\cos {}^3\mathrm{x}=\sin {}^2\mathrm{x}+\cos {}^2\mathrm{x}-\frac{1}{2}\sin {}^2\mathrm{x}$$$$\sin {}^3\mathrm{x}+\cos {}^3\mathrm{x}=\frac{1}{2}\sin {}^2\mathrm{x}+\cos {}^2\mathrm{x}$$$$\cos {}^3\mathrm{x}-\cos {}^2\mathrm{x}=\frac{1}{2}\sin {}^2\mathrm{x}-\sin {}^3\mathrm{x}$$$$\cos {}^2\mathrm{x}(\cos \mathrm{x}-1)=\sin {}^2\mathrm{x}(\frac{1}{2}-\sin \mathrm{x})$$$$\cos {}^2\mathrm{x}(-2\sin {}^2\frac{1}{2}\mathrm{x})=\sin {}^2\mathrm{x}(\frac{1}{2}-2\sin \frac{1}{2}\mathrm{xcos}\frac{1}{2}\mathrm{x})$$$$\mathrm{let}\tan \frac{\mathrm{x}}{2}=\mathrm{t}\Rightarrow \tan \mathrm{x}=\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}$$$$\Rightarrow -2\left(\frac{{\mathrm{t}}^2}{1+{\mathrm{t}}^2}\right)=\left(\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}\right)(\frac{1}{2}-\frac{2\mathrm{t}}{1+{\mathrm{t}}^2})$$$$\Rightarrow \frac{-{2\mathrm{t}}^2}{1+{\mathrm{t}}^2}=\left(\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}\right)\left(\frac{1+{\mathrm{t}}^2-4\mathrm{t}}{2(1+{\mathrm{t}}^2)}\right)$$$$\Rightarrow \left(\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\right)[\frac{{\mathrm{t}}^2-4\mathrm{t}+1}{2-{2\mathrm{t}}^2}+\mathrm{t}]=0$$$$\Rightarrow \left(\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\right)\left[\frac{{\mathrm{t}}^2-4\mathrm{t}+1+2\mathrm{t}-{2\mathrm{t}}^3}{2-{2\mathrm{t}}^2}\right]=0$$$$\Rightarrow \left(\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\right)\left(\frac{{2\mathrm{t}}^3-{\mathrm{t}}^2+2\mathrm{t}-1}{2-{2\mathrm{t}}^2}\right)=0$$$$\mathrm{we}\mathrm{get}\mathrm{t}=0$$$$\Rightarrow \tan \frac{\mathrm{x}}{2}=0\to \frac{\mathrm{x}}{2}=0+\mathrm{k}\pi$$$$\Rightarrow \mathrm{x}=2\mathrm{k}\pi ;\mathrm{k}\in \mathbb{Z}$$$$\mathrm{for}{2\mathrm{t}}^3-{\mathrm{t}}^2+2\mathrm{t}-1=0$$$$\mathrm{test}\mathrm{t}=\frac{1}{2}\to \frac{1}{4}-\frac{1}{4}+1-1=0\left(\mathrm{valid}\right)$$$$\Rightarrow \mathrm{factorise}(2\mathrm{t}-1)({\mathrm{t}}^2+1)=0$$$$\Rightarrow \tan \frac{\mathrm{x}}{2}=\frac{1}{2};\frac{\mathrm{x}}{2}=\arctan \left(\frac{1}{2}\right)+\mathrm{k}\pi$$$$\Rightarrow \mathrm{x}=2\arctan \left(\frac{1}{2}\right)+2\mathrm{k}\pi$$

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