Question Number 144333 by gsk2684 last updated on 24/Jun/21
![find the number of solutions of 1+ sin x.sin^2 (x/2)=0 in [−Π Π]](https://www.tinkutara.com/question/Q144333.png)
$$\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\mathrm{1}+\:\mathrm{sin}\:\mathrm{x}.\mathrm{sin}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}=\mathrm{0}\:\mathrm{in}\:\left[−\Pi\:\Pi\right] \\ $$
Commented by MJS_new last updated on 25/Jun/21
$$\mathrm{no}\:\mathrm{solution}: \\ $$$$\mathrm{1}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\leqslant\mathrm{1}+\mathrm{sin}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\:\leqslant\mathrm{1}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$
Commented by MJS_new last updated on 25/Jun/21
![let t=tan (x/2) ⇒ sin x =((2t)/(t^2 +1))∧sin (x/2) =(t/( (√(t^2 +1)))) 1+sin x sin^2 (x/2)=((t^4 +2t^3 +2t^2 +1)/((t^2 +1)^2 )) (d/dt)[((t^4 +2t^3 +2t^2 +1)/((t^2 +1)^2 ))]=0 −((2t^2 (t^2 −3))/((t^2 +1)^3 ))=0 ⇒ t=0∨t=±(√3) ⇒ 1−((3(√3))/8)≤((t^4 +2t^3 +2t^2 +1)/((t^2 +1)^2 ))≤1+((3(√3))/8)](https://www.tinkutara.com/question/Q144378.png)
$$\mathrm{let}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Rightarrow\:\mathrm{sin}\:{x}\:=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\wedge\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{1}+\mathrm{sin}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}=\frac{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dt}}\left[\frac{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right]=\mathrm{0} \\ $$$$−\frac{\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{3}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0}\:\Rightarrow\:{t}=\mathrm{0}\vee{t}=\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{1}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\leqslant\frac{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\leqslant\mathrm{1}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$
Commented by gsk2684 last updated on 26/Jun/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{MJS}. \\ $$$$ \\ $$