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Question Number 14267 by Tinkutara last updated on 30/May/17

$$\mathrm{If}{\sin }^{3}x+{\cos }^{3}x+3\sin x\cos x-1=0,$$$$\mathrm{then}\mathrm{find}x.$$

Answered by linkelly0615 last updated on 30/May/17

$$set$$$$\{\begin{array}{l}A=sinx+cosx\\ B=sinx-cosx\end{array}$$$$\Rightarrow$$$${\sin }^{3}x+{\cos }^{3}x+3\sin x\cos x-1=0$$$$\Rightarrow A(A^2-2sinxcosx)-3Asinxcosx+3sinxcosx-1=0$$$$\Rightarrow (1-A)[3sinxcosx-(1+A+A^2)]=0$$$$\Rightarrow (1-A)[\frac{3}{4}(A^2-B^2)-(1+A+A^2)]=0$$$$\Rightarrow (A-1)\left\{\frac{1}{4}[{(A+2)}^2+3B^2]\right\}=0$$$$\Rightarrow A=1or\{\begin{array}{l}A=(-2)\\ B=0\end{array}$$$$!!BUT!!$$$$\because A=sinx+cosx=\sqrt{2}\sin (x+\frac{\pi }{4})$$$$\therefore \mid A\mid ⩽\sqrt{2}\Rightarrow A{won}^{\prime }tbe(-2)$$$$$$$$\Rightarrow so,A=1$$$$\{sorry,{that}^{\prime }smyfault.\}$$$$\Rightarrow sinx+cosx=1$$$$[sinx+cosx=\sqrt{}2(\frac{sinx}{\sqrt{2}}+\frac{cosx}{\sqrt{2}})=\sqrt{2}sin(x+\frac{\pi }{4})]$$$$\Rightarrow sin(x+\pi /4)=1/\sqrt{2}$$$$\Rightarrow x+\pi /4=2k\pi +\pi /4or2k\pi +3\pi /4,k\in \mathbb{Z}$$$$\Rightarrow x=2k\pi or(2k+\frac{1}{2})\pi ,k\in \mathbb{Z}$$

Commented by Tinkutara last updated on 30/May/17

$$\mathrm{Can}\mathrm{you}\mathrm{explain}\mathrm{the}\mathrm{last}\mathrm{line}?\mathrm{My}\mathrm{doubt}$$$$\mathrm{is}\sin x=0\mathrm{implies}x=n\pi \mathrm{and}$$$$\cos x=0\mathrm{implies}x=(2n+1)\frac{\pi }{2}.$$$$\mathrm{Then}\mathrm{why}\mathrm{you}\mathrm{have}\mathrm{written}x=2k\pi$$$$\mathrm{or}(2k+\frac{1}{2})\pi ?$$

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