Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 200873 by cortano12 last updated on 25/Nov/23

Commented by witcher3 last updated on 26/Nov/23

$$\tan \left(\frac{3\mathrm{x}}{2}\right)\in [-\sqrt{3},\sqrt{3}]$$$$\mathrm{tg}(3.\frac{\mathrm{x}}{2})=\frac{3\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right)-{\mathrm{tg}}^3\left(\frac{\mathrm{x}}{2}\right)}{1-{3\mathrm{tg}}^2\left(\frac{\mathrm{x}}{2}\right)}$$$$\sin \left(\mathrm{x}\right)\sqrt{3-{\tan }^2\left(\frac{3\mathrm{x}}{2}\right)}=2+\cos \left(\mathrm{x}\right)$$$$\iff \sqrt{3-{\left[\frac{3\tan \left(\frac{\mathrm{x}}{2}\right)-{\tan }^3\left(\frac{\mathrm{x}}{2}\right)}{1-{3\tan }^2\left(\frac{\mathrm{x}}{2}\right)}\right]}^2}=\frac{{3\cos }^2\left(\frac{\mathrm{x}}{2}\right)+{\sin }^2\left(\frac{\mathrm{x}}{2}\right)}{2\sin \left(\frac{\mathrm{x}}{2}\right)\cos \left(\frac{\mathrm{x}}{2}\right)}=\frac{3+{\mathrm{tg}}^2\left(\frac{\mathrm{x}}{2}\right)}{2\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right)}$$$$\mathrm{a}=\tan \left(\frac{\mathrm{x}}{2}\right)$$$$\iff \sqrt{3-{\left[\frac{(3\mathrm{a}-{\mathrm{a}}^3)}{{(1-{3\mathrm{a}}^2)}^2}\right]}^2}=\frac{3+{\mathrm{a}}^2}{2\mathrm{a}}$$$$\Rightarrow \frac{3({9\mathrm{a}}^4-{6\mathrm{a}}^2+1)-({9\mathrm{a}}^2+{\mathrm{a}}^6-{6\mathrm{a}}^4)}{{(1-{3\mathrm{a}}^2)}^2}=\frac{({\mathrm{a}}^4+{6\mathrm{a}}^2+9)}{{4\mathrm{a}}^2}$$$${\mathrm{a}}^2=\mathrm{y}$$$$\iff \frac{-{\mathrm{y}}^3+{33\mathrm{y}}^2-27\mathrm{y}+3}{{(1-3\mathrm{y})}^2}=\frac{{\mathrm{y}}^2+6\mathrm{y}+9}{4\mathrm{y}}$$$$4\mathrm{y}(-{\mathrm{y}}^3+{33\mathrm{y}}^2-27\mathrm{y}+3)=({9\mathrm{y}}^2-6\mathrm{y}+1)({\mathrm{y}}^2+6\mathrm{y}+9)$$$${9\mathrm{y}}^4+{48\mathrm{y}}^3+{46\mathrm{y}}^2-48\mathrm{y}+9=-{4\mathrm{y}}^4+{132\mathrm{y}}^3-{108\mathrm{y}}^2+12\mathrm{y}$$$${13\mathrm{y}}^4-{84\mathrm{y}}^3+{154\mathrm{y}}^2-60\mathrm{y}+9=0...\mathrm{P}$$$$\mathrm{In}\mathbb{Z}/3\mathbb{Z}\Rightarrow {2\mathrm{y}}^2\equiv 0\left[3\right]\Rightarrow 3\mid \mathrm{y}$$$$\mathrm{y}=3\mathrm{is}\mathrm{solution}\mathrm{of}\mathrm{P}$$$${13\mathrm{y}}^4-{84\mathrm{y}}^3+{154\mathrm{y}}^2-60\mathrm{y}+9$$$$={(\mathrm{y}-3)}^2({13\mathrm{y}}^2-6\mathrm{y}+1)$$$$\mathrm{only}\mathrm{one}\mathrm{solution}\mathrm{over}\mathbb{R}$$$$\mathrm{a}=\underset{-}{+}\sqrt{3}$$$$\tan \left(\frac{\mathrm{x}}{2}\right)=\sqrt{3}\Rightarrow \frac{\mathrm{x}}{2}={\tan }^{-1}\left(\sqrt{3}\right)\Rightarrow \mathrm{x}=\frac{2\pi }{3}+2\mathrm{k}\pi ..\mathrm{worck}$$$$\tan \left(\frac{\mathrm{x}}{2}\right)=-\sqrt{3}\Rightarrow \mathrm{x}=-\frac{2\pi }{3}+2\mathrm{k}\pi ...\mathrm{not}\mathrm{worcking}$$$$$$$$$$

Answered by Frix last updated on 26/Nov/23

$$c=\cos x$$$$\sqrt{(1-c)(1+c)}\sqrt{3-\frac{(1-c){(1+2c)}^2}{(1+c){(1-2c)}^2}}-c=2$$$$c\ne -1\wedge c\ne \frac{1}{2}$$$$\sqrt{1-c}\sqrt{2(8c^3-6c+1)}=(2+c)\mid 1-2c\mid$$$$c^4-\frac{c^3}{5}-\frac{11c^2}{20}+\frac{c}{10}+\frac{1}{10}=0$$$${(c+\frac{1}{2})}^2(c^2-\frac{6c}{5}+\frac{2}{5})=0$$$$c=-\frac{1}{2}$$$$\cos x=-\frac{1}{2}$$$$x=2n\pi \pm \frac{2\pi }{3}$$$$\mathrm{Testing}\Rightarrow x=2n\pi +\frac{2\pi }{3}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com