Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 48434 by behi83417@gmail.com last updated on 23/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Nov/18

$$tanx=t$$$$t(2+\sqrt{3}\times \frac{3t-t^3}{1-3t^2})=\frac{3t-t^3}{1-3t^2}$$$$2+\sqrt{3}\times \frac{3t-t^3}{1-3t^2}=\frac{3-t^2}{1-3t^2}(t\ne 0)$$$$\frac{3\sqrt{3}t-\sqrt{3}{t}^3}{1-3t^2}=\frac{3-t^2}{1-3t^2}-2$$$$3\sqrt{3}t-\sqrt{3}{t}^3=3-t^2-2+6t^2$$$$-\sqrt{3}{t}^3-5t^2+3\sqrt{3}t-1=0$$$$\sqrt{3}{t}^3+5t^2-3\sqrt{3}t+1=0$$$$solving....$$$$1)\sqrt{3}t(t^2-3)+5t^{2}shouldbeequalsto-1$$$$2)5t^{2}is+veso\sqrt{3}t(t^2-3)<0$$$$tomake\sqrt{3}t(t^2-3)+5t^2-veandequalsfto(-1)$$$$3)criticalvalueof\sqrt{3}t(t^2-3)are$$$$0,-\sqrt{3}and+\sqrt{3}$$$$f\left(t\right)=\sqrt{3}t(t^2-3)+5t^2+1$$$$t\ne 0t\ne \sqrt{3}t\ne -\sqrt{3}$$$$f\left(0\right)>0+ve$$$$f\left(1\right)>0+ve$$$$f(0.5)<0$$$$f(0.6)>0$$$$so0.6>tanx>0.5$$$$plscheck...$$$$$$

Commented by behi83417@gmail.com last updated on 24/Nov/18

$$thankyouverymuchsir.$$

Answered by MJS last updated on 23/Nov/18

$$x=\arctan t$$$$\frac{\sqrt{3}{t}^4+5t^3-3\sqrt{3}{t}^2+t}{3t^2-1}=0$$$$\Rightarrow 3t^2-1\ne 0\Rightarrow t\ne \pm \frac{\sqrt{3}}{3}$$$$t(\sqrt{3}{t}^3+5t^2-3\sqrt{3}t+1)=0$$$$\Rightarrow t_1=0\Rightarrow x_1=0$$$$t^3+\frac{5\sqrt{3}}{3}{t}^2-3t+\frac{\sqrt{3}}{3}=0$$$$\mathrm{trying}t=\pm \frac{\sqrt{3}}{3}\Rightarrow t_2=\frac{\sqrt{3}}{3}\mathrm{but}t\ne \frac{\sqrt{3}}{3}\Rightarrow \mathrm{no}\mathrm{solution}\mathrm{of}\mathrm{given}\mathrm{eq}$$$$t^2+2\sqrt{3}t-1=0$$$$\Rightarrow t_3=-2-\sqrt{3};t_4=2-\sqrt{3}$$$$\Rightarrow x_3=-\frac{5\pi }{12};x_4=\frac{\pi }{12}$$$$\mathrm{answer}\mathrm{is}$$$$(x=z\pi \vee x=-\frac{5\pi }{12}+z\pi \vee x=\frac{\pi }{12}+z\pi )\wedge z\in \mathbb{Z}$$

Commented by behi83417@gmail.com last updated on 24/Nov/18

$$thanksinadvancesir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com