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Question Number 68666 by ahmadshah last updated on 14/Sep/19

Commented by mr W last updated on 14/Sep/19

Commented by mr W last updated on 14/Sep/19

$$\alpha =55+25=80°$$$$x=\alpha +40=80+40=120°$$

Answered by ahmadshah last updated on 14/Sep/19

Commented by ahmadshah last updated on 14/Sep/19

Commented by Prithwish sen last updated on 15/Sep/19

$$\mathrm{You}\mathrm{have}\mathrm{a}\mathrm{creative}\mathrm{mind}.\mathrm{Thanks}\mathrm{rasheed}\mathrm{Sir}.$$

Commented by Prithwish sen last updated on 16/Sep/19

$$({\mathrm{x}}^2+1)-3\sqrt{{\mathrm{x}}^3+1}+2\sqrt{{\mathrm{x}}^3+1}-6\mathrm{x}=0$$$$\sqrt{{\mathrm{x}}^2+1}(\sqrt{{\mathrm{x}}^2+1}-3\sqrt{\mathrm{x}})+2\sqrt{\mathrm{x}}(\sqrt{{\mathrm{x}}^2+1}-3\sqrt{\mathrm{x}})=0$$$$(\sqrt{{\mathrm{x}}^2+1}+2\sqrt{\mathrm{x}})(\sqrt{{\mathrm{x}}^2+1}-3\sqrt{\mathrm{x}})=0$$$$\therefore \mathbf{either}\mathbf{or}$$$$\sqrt{{\mathbf{x}}^2+1}+2\sqrt{\mathbf{x}}=0\sqrt{{\mathbf{x}}^2+1}-3\sqrt{\mathbf{x}}=0$$$${\mathbf{x}}^2-4\mathbf{x}+1=0{\mathbf{x}}^2-9\mathbf{x}+1=0$$$$\mathbf{x}=\frac{4\pm \sqrt{16-4}}{2}\mathbf{x}=\frac{9\pm \sqrt{81-4}}{2}$$$$\therefore \mathrm{x}=2\pm \sqrt{3},\frac{9\pm \sqrt{77}}{2}\mathbf{please}\mathbf{check}.$$

Commented by Rasheed.Sindhi last updated on 15/Sep/19

$${\mathbb{A}}^{\mathbf{n}}{\mathbb{E}}^{\mathbf{asier}}{\mathbb{A}}^{\mathbf{pproach}}$$$${\mathrm{x}}^2+1-\sqrt{{\mathrm{x}}^3+\mathrm{x}}=6\mathrm{x}$$$$\mathrm{Dividing}\mathrm{by}{\mathrm{x}}^2+1$$$$1-\frac{\sqrt{\mathrm{x}}\sqrt{{\mathrm{x}}^2+1}}{{\mathrm{x}}^2+1}=6\left(\frac{\mathrm{x}}{{\mathrm{x}}^2+1}\right)$$$$1-\frac{\sqrt{\mathrm{x}}\sqrt{{\mathrm{x}}^2+1}}{{\left(\sqrt{{\mathrm{x}}^2+1}\right)}^2}=6{\left(\frac{\sqrt{\mathrm{x}}}{\sqrt{{\mathrm{x}}^2+1}}\right)}^2$$$$1-\frac{\sqrt{\mathrm{x}}}{\sqrt{{\mathrm{x}}^2+1}}=6{\left(\frac{\sqrt{\mathrm{x}}}{\sqrt{{\mathrm{x}}^2+1}}\right)}^2$$$$\mathrm{Let}\frac{\sqrt{\mathrm{x}}}{\sqrt{{\mathrm{x}}^2+1}}=\mathrm{y}$$$$1-\mathrm{y}={6\mathrm{y}}^2$$$${6\mathrm{y}}^2+\mathrm{y}-1=0$$$$(2\mathrm{y}+1)(3\mathrm{y}-1)=0$$$$\mathrm{y}=-\frac{1}{2}\mid \mathrm{y}=\frac{1}{3}$$$$\frac{\sqrt{\mathrm{x}}}{\sqrt{{\mathrm{x}}^2+1}}=-\frac{1}{2}\mid \frac{\sqrt{\mathrm{x}}}{\sqrt{{\mathrm{x}}^2+1}}=\frac{1}{3}$$$$\frac{\mathrm{x}}{{\mathrm{x}}^2+1}=\frac{1}{4}\mid \frac{\mathrm{x}}{{\mathrm{x}}^2+1}=\frac{1}{9}$$$${\mathrm{x}}^2-4\mathrm{x}+1=0\mid {\mathrm{x}}^2-9\mathrm{x}+1=0$$

Commented by Prithwish sen last updated on 15/Sep/19

$$\mathrm{Excellent}\mathrm{Sir}.$$

Commented by Rasheed.Sindhi last updated on 15/Sep/19

$$\mathbf{Sir},\mathrm{actually}\mathrm{this}\mathrm{solution}\mathrm{raised}\mathrm{in}$$$$\mathrm{my}\mathrm{mind}\mathrm{after}\mathrm{seeing}\mathrm{your}\mathrm{solution}.$$$$\mathrm{I}\mathrm{like}\mathrm{your}\mathrm{solution}!$$

Commented by mind is power last updated on 15/Sep/19

$$x=tg\left(t\right)..x⩾0\Rightarrow t\in [0,\frac{\pi }{2}]$$$$\Rightarrow \frac{1}{{cos}^2\left(t\right)}-\sqrt{\frac{sin\left(t\right)}{{cos}^3\left(t\right)}}=6\frac{sin\left(t\right)}{cos\left(t\right)}$$$$$$$$\Rightarrow 1-\sqrt{sin\left(t\right)cos\left(t\right)}=6sin\left(t\right)cos\left(t\right)$$$$\Rightarrow 1-\sqrt{\frac{sin\left(2t\right)}{2}}=3sin\left(2t\right)$$$$\Rightarrow \sqrt{sin\left(2t\right)}solutionof1-\frac{X}{\sqrt{2}}=3X^2$$

Commented by Rasheed.Sindhi last updated on 15/Sep/19

$$Sirmind-is-power,Niceideaof$$$$connectingthealgebraicproblem$$$$totrigonom\mathrm{etry}!$$

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