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Question Number 148797 by kameda last updated on 31/Jul/21

Answered by puissant last updated on 31/Jul/21

$$1)$$$$\sin \alpha =\frac{\mathrm{AB}}{\mathrm{OA}}\Rightarrow \sin \alpha =\frac{\mathrm{AB}}{1}\mathrm{car}{\mathrm{c}}^{\prime }\mathrm{est}\mathrm{un}$$$$\mathrm{cercle}\mathrm{de}\mathrm{rayon}1..$$$$\Rightarrow \mathrm{AB}=\sin \left(\mathrm{x}\right)\mathrm{car}\alpha \mathrm{est}\mathrm{represent}\acute{\mathrm{e}}\mathrm{par}\mathrm{x}.$$$${\mathrm{A}}_{\mathrm{\Delta }}=\frac{\mathrm{AB}\times \mathrm{h}}{2}=\frac{\mathrm{AB}}{2}=\frac{\sin \left(\mathrm{x}\right)}{2}..\mathrm{car}\mathrm{h}=1$$$${\mathrm{D}}^{\prime }\mathrm{autre}\mathrm{part},{\mathrm{l}}^{\prime }\mathrm{air}\mathrm{de}{\mathrm{l}}^{\prime }\mathrm{arc}\mathrm{AB}\mathrm{est}:$$$${\mathrm{A}}_{{\mathrm{l}}^{\prime }\mathrm{arc}}+{\mathrm{A}}_{△}=\frac{\alpha }{2}=\frac{\mathrm{x}}{2}$$$$\Rightarrow {\mathrm{A}}_{{\mathrm{l}}^{\prime }\mathrm{arc}}=\frac{\mathrm{x}}{2}-{\mathrm{A}}_{△}=\frac{\mathrm{x}}{2}-\frac{\sin \left(\mathrm{x}\right)}{2}..$$$$\mathrm{Alors},\mathrm{pour}\mathrm{que}{\mathrm{A}}_{{\mathrm{l}}^{\prime }\mathrm{arc}}={\mathrm{A}}_{△},\mathrm{il}\mathrm{faut}\mathrm{que}:$$$${\mathrm{A}}_{△}-{\mathrm{A}}_{{\mathrm{l}}^{\prime }\mathrm{arc}}=0$$$$\Rightarrow \frac{\sin \left(\mathrm{x}\right)}{2}-\frac{\mathrm{x}}{2}+\frac{\sin \left(\mathrm{x}\right)}{2}=0$$$$\Rightarrow \sin \left(\mathrm{x}\right)-\frac{\mathrm{x}}{2}=0$$$$\mathrm{posons}\mathrm{A}\left(\mathrm{x}\right)=\sin \left(\mathrm{x}\right)-\frac{\mathrm{x}}{2}..$$$$\mathrm{etudier}\mathrm{la}\mathrm{fonction}\mathrm{et}\mathrm{trouver}{\mathrm{l}}^{\prime }\mathrm{intervalle}$$$$\mathrm{ou}\mathrm{la}\mathrm{variable}\mathrm{se}\mathrm{trouve}..$$$$2)$$$$....\mathrm{Trivial}..$$

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