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Question Number 72492 by petrochengula last updated on 29/Oct/19

$$\frac{1}{\mathrm{cosx}}+\frac{1}{\mathrm{sinx}}=8$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}$$

Commented by petrochengula last updated on 29/Oct/19

$$\mathrm{help}\mathrm{please}$$

Answered by behi83417@gmail.com last updated on 29/Oct/19

$$\mathrm{sinx}+\mathrm{cosx}=8\mathrm{sinx}.\mathrm{cosx}$$$${(\mathrm{sinx}+\mathrm{cosx})}^2={(8\mathrm{sinx}.\mathrm{cosx})}^2$$$$1+2\mathrm{sinx}.\mathrm{cosx}={64\sin }^2{\mathrm{xcos}}^2\mathrm{x}[\mathrm{let}:\mathrm{sinxcosx}=\mathrm{y}]$$$${65\mathrm{y}}^2={\mathrm{y}}^2+2\mathrm{y}+1={(\mathrm{y}+1)}^2$$$$\Rightarrow \sqrt{65}\mathrm{y}=\pm (2\mathrm{y}+1)\Rightarrow \{\begin{array}{l}\mathrm{y}=\frac{1}{\sqrt{65}-2}\\ \mathrm{y}=\frac{-1}{\sqrt{65}+2}\end{array}$$$$1)\mathrm{sinx}.\mathrm{cosx}=\frac{1}{\sqrt{65}-2}\Rightarrow \sin 2\mathrm{x}=\frac{2}{\sqrt{65}-2}$$$$\Rightarrow \mathrm{x}=\mathrm{k}\pi \pm \frac{1}{2}{\sin }^{-1}\left[\frac{2}{\sqrt{65}-2}\right](\mathrm{k}\in \mathrm{z})$$$$2)\mathrm{sinx}.\mathrm{cosx}=\frac{-1}{\sqrt{65}+2}\Rightarrow \sin 2\mathrm{x}=\frac{-2}{\sqrt{65}+2}$$$$\Rightarrow \mathrm{x}=\mathrm{l}\pi \pm \frac{1}{2}{\sin }^{-1}\left[\frac{-2}{\sqrt{65}+2}\right](\mathrm{l}\in \mathrm{z})$$

Commented by petrochengula last updated on 29/Oct/19

$$\mathrm{thanks}$$

Answered by Tanmay chaudhury last updated on 29/Oct/19

$$sinx+cosx=4\left(2sinxcosx\right)$$$$\sqrt{1+sin2x}=4sin2x$$$$1+a=16a^2\to 16a^2-a-1=0$$$$a=\frac{1\pm \sqrt{1-4.16.(-1)}}{2\times 16}=\frac{1\pm \sqrt{65}}{32}$$$$\sqrt{65}\approx 8.06$$$$a=\frac{9.06}{32},\frac{-7.06}{32}$$$$a\approx 0.28,-0.22$$$$sin2x=0.28=sin(0.284)[noteangleinradian$$$$2x=0.284\to x=0.142$$$$tookhelpofcalculator$$

Commented by petrochengula last updated on 29/Oct/19

$$\mathrm{thanks}$$

Commented by Tanmay chaudhury last updated on 29/Oct/19

$$mostwelcome...$$

Answered by MJS last updated on 29/Oct/19

$$x=2\arctan t\Rightarrow \cos x=\frac{1-t^2}{1+t^2};\sin x=\frac{2t}{1+t^2}$$$$\frac{1+t^2}{1-t^2}+\frac{1+t^2}{2t}=8$$$$\mathrm{transforming}\Rightarrow$$$$t^4-18t^3+14t-1=0$$$$\mathrm{we}\mathrm{can}\mathrm{exactly}\mathrm{solve}\mathrm{this}$$$$(t^2-(9+\sqrt{65})t-(8+\sqrt{65}))(t^2-(9-\sqrt{65})t-(8r-\sqrt{65}))=0$$$$t_{1,2}=\frac{9+\sqrt{65}}{2}\pm \frac{\sqrt{178+22\sqrt{65}}}{2}$$$$t_{3,4}=\frac{9-\sqrt{65}}{2}\pm \frac{\sqrt{178-22\sqrt{65}}}{2}$$$$x_i=2\pi n+2\arctan t\mathrm{with}n\in \mathbb{Z}$$$$\Rightarrow$$$$x_1\approx -1.45953+2\pi n$$$$x_2\approx 3.03033+2\pi n$$$$x_3\approx .143562+2\pi n$$$$x_4\approx 1.42723+2\pi n$$

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