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Question Number 144333 by gsk2684 last updated on 24/Jun/21

$$\mathrm{find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{solutions}\mathrm{of}$$$$1+\sin \mathrm{x}.{\sin }^2\frac{\mathrm{x}}{2}=0\mathrm{in}[-\mathrm{\Pi }\mathrm{\Pi }]$$

Commented by MJS_new last updated on 25/Jun/21

$$\mathrm{no}\mathrm{solution}:$$$$1-\frac{3\sqrt{3}}{8}⩽1+\sin x{\sin }^2\frac{x}{2}⩽1+\frac{3\sqrt{3}}{8}$$

Commented by MJS_new last updated on 25/Jun/21

$$\mathrm{let}t=\tan \frac{x}{2}\Rightarrow \sin x=\frac{2t}{t^2+1}\wedge \sin \frac{x}{2}=\frac{t}{\sqrt{t^2+1}}$$$$1+\sin x{\sin }^2\frac{x}{2}=\frac{t^4+2t^3+2t^2+1}{{(t^2+1)}^2}$$$$\frac{d}{dt}\left[\frac{t^4+2t^3+2t^2+1}{{(t^2+1)}^2}\right]=0$$$$-\frac{2t^2(t^2-3)}{{(t^2+1)}^3}=0\Rightarrow t=0\vee t=\pm \sqrt{3}$$$$\Rightarrow 1-\frac{3\sqrt{3}}{8}⩽\frac{t^4+2t^3+2t^2+1}{{(t^2+1)}^2}⩽1+\frac{3\sqrt{3}}{8}$$

Commented by gsk2684 last updated on 26/Jun/21

$$\mathrm{thank}\mathrm{you}\mathrm{MJS}.$$$$$$

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