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Question Number 14078 by Tinkutara last updated on 27/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/May/17

$$4{cos}^{3}x-3cosx-\sqrt{3}sinx.cosx+{cos}^{2}x-\frac{1}{2}=-2$$$$8{cos}^{3}x+2{cos}^{2}x-2\sqrt{3}sinx.cosx-6cosx+3=0$$$$cosx=t\Rightarrow$$$$8t^3+2t^2-6t+3=2\sqrt{3}t\sqrt{1-t^2}$$$$64t^6+4t^4+36t^2+9+32t^5-96t^4+48t^3-24t^3+12t^2-36t=12t^2-12t^4$$$$\Rightarrow 64t^6+32t^5-80t^4+24t^3+36t^2-36t+9=0$$$$\Rightarrow t=cosx=\frac{1}{2}\Rightarrow x=2k\pi \pm \frac{\pi }{3}.◼$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/May/17

Commented by ajfour last updated on 28/May/17

$$\bullet \sin [2(2k\pi -\frac{\pi }{3})-\frac{7\pi }{6}]$$$$=\sin (4k\pi -\frac{11\pi }{6})=\sin \frac{\pi }{6}\ne -1.$$$$\bullet \sin [2(2k\pi +\frac{\pi }{3})-\frac{7\pi }{6}]$$$$=\sin (4k\pi -\frac{\pi }{2})=-1.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 28/May/17

$$cos(-x)=cox\left(x\right)=\frac{1}{2}$$

Answered by ajfour last updated on 27/May/17

$$x=2n\pi +\frac{\pi }{3}.$$

Commented by Tinkutara last updated on 27/May/17

$$\mathrm{Can}\mathrm{you}\mathrm{explain}\mathrm{the}\mathrm{solution}\mathrm{please}?$$

Answered by ajfour last updated on 27/May/17

$$aboveequationimpliesthat$$$$\cos 3x=-1and\sin (2x-\frac{7\pi }{6})=-1$$$$\Rightarrow 3x=\pi +2k\pi ;k\in Z$$$$x=\frac{\pi }{3}+k\left(\frac{2\pi }{3}\right).....\left(i\right)$$$$also2x-\frac{7\pi }{6}=2n\pi -\frac{\pi }{2}$$$$\Rightarrow x=\frac{\pi }{3}+n\pi .....\left(ii\right)$$$$foreqns.\left(i\right)and\left(ii\right)tobe$$$$simultaneouslytrue:$$$$\mathit{x}=2\mathit{n}\mathit{\pi }+\mathit{\pi }/3.$$

Commented by Tinkutara last updated on 27/May/17

$$\mathrm{But}\mathrm{since}\cos 3x=-1=\cos \pi ,$$$$\therefore 3x=2n\pi \pm \pi$$$$\mathrm{and}\mathrm{similarly}\sin x=\sin y\mathrm{implies}$$$$x=n\pi +{(-1)}^{n}y$$$$\mathrm{Why}\mathrm{you}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{use}\mathrm{these}\mathrm{conditions}?$$

Commented by ajfour last updated on 27/May/17

Commented by mrW1 last updated on 27/May/17

$$\cos \left(a\right)⩾-1$$$$\sin \left(b\right)⩾-1$$$$\Rightarrow \cos \left(a\right)+\sin \left(b\right)⩾-2$$$$‘‘{=}^{″}isvalidonlyif$$$$\cos \left(a\right)=-1$$$$\sin \left(b\right)=-1$$

Commented by RasheedSindhi last updated on 27/May/17

$$aboveequation\stackrel{?}{implies}that$$$$\cos 3x=-1and\sin (2x-\frac{7\pi }{6})=-1$$$$-------------$$$$\mathrm{Why}\mathrm{the}\mathrm{above}\mathrm{eq}\mathrm{implies}?$$$$\mathrm{a}+\mathrm{b}=\mathrm{c}[=\mathrm{d}+\mathrm{e}\left(\mathrm{say}\right)]$$$$\Rightarrow \mathrm{a}=\mathrm{d}\wedge \mathrm{b}=\mathrm{e}?$$

Answered by mrW1 last updated on 27/May/17

$$since-2⩽\cos \left(a\right)+\sin \left(b\right)⩽2$$$$for\cos 3x+\sin (2x-\frac{7\pi }{6})=-2$$$$\Rightarrow \cos 3x=-1$$$$\Rightarrow \sin (2x-\frac{7\pi }{6})=-1$$$$\Rightarrow 3x=(2n+1)\pi ,n\in \mathbb{Z}$$$$x=\frac{2n+1}{3}\pi =\frac{2n\pi }{3}+\frac{\pi }{3}$$$$\Rightarrow 2x-\frac{7\pi }{6}=2m\pi -\frac{\pi }{2}$$$$\Rightarrow x=m\pi +\frac{\pi }{3}$$$$\frac{2n}{3}=m$$$$\Rightarrow n=3k,k\in \mathbb{Z}$$$$$$$$Solutionis$$$$x=2k\pi +\frac{\pi }{3}$$

Commented by Tinkutara last updated on 28/May/17

$$\mathrm{But}\mathrm{if}\cos x=\cos y,\mathrm{we}\mathrm{should}\mathrm{use}$$$$x=2n\pi \pm y\mathrm{and}\mathrm{for}\sin x=\sin y,$$$$\mathrm{it}\mathrm{should}\mathrm{be}x=n\pi +{(-1)}^{n}y.$$

Commented by ajfour last updated on 28/May/17

$$whenwehavespecialvalues$$$$ofysuchasy=n\left(\frac{\pi }{2}\right)we$$$$shouldpreferwritinginthat$$$$manner(youareasking$$$$justificationfor).$$$$$$$$if\cos x=-1$$$$x=2k\pi +\pi$$$$orx=2n\pi \pm \pi areequivalent$$$$sinceforn=0,youhave$$$$x=-\pi ,\pi$$$$bothcanbeobtainedwithk=-1$$$$andk=0$$$$forn=1,x=\pi ,3\pi$$$$samevaluesobtainedwithk=0,$$$$k=1$$$$whatiscommontobothexpressions$$$$isthatx=\left(oddinteger\right)\times \pi$$$$thisonlyisnecessaryfor$$$$\cos x=-1.$$$$Alongsimilarlines,$$$$if\sin x=-1$$$$x=2k\pi -\frac{\pi }{2}$$$$orx=n\pi +{(-1)}^n(-\frac{\pi }{2})would$$$$againbeequivalentrepresentations.$$

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