Question Number 80896 by mathocean1 last updated on 07/Feb/20
![α is a real ∈ ]0;(π/2)[. we give this (E_α ):2x^2 −2x(√2)(cosα)+cos2α=0 1. show that Δ=8sin^2 x i showed it. 2.Solve E_α in R. please help me for this question.](Q80896.png)
$$\left.\alpha\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\in\:\right]\mathrm{0};\frac{\pi}{\mathrm{2}}\left[.\:\mathrm{we}\:\mathrm{give}\:\mathrm{this}\:\right. \\ $$$$\left(\mathrm{E}_{\alpha} \right):\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\sqrt{\mathrm{2}}\left({cos}\alpha\right)+\mathrm{cos2}\alpha=\mathrm{0} \\ $$$$\mathrm{1}.\:\mathrm{show}\:\mathrm{that}\:\Delta=\mathrm{8sin}^{\mathrm{2}} {x} \\ $$$${i}\:{showed}\:{it}. \\ $$$$\mathrm{2}.{S}\mathrm{olve}\:\mathrm{E}_{\alpha} \:\mathrm{in}\:\mathbb{R}. \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{for}\:\mathrm{this}\:\mathrm{question}. \\ $$
Answered by behi83417@gmail.com last updated on 07/Feb/20
![△′=[(√2).cosα]^2 −2cos2α= =2cos^2 α−2(cos^2 α−sin^2 α)=2sin^2 α ⇒x=((−b′±(√△))/a)=(((√2).cos𝛂±(√2)sin𝛂)/2)= =((√2)/2)cos𝛂±((√2)/2)sin𝛂=cos((𝛑/4)±𝛂) .](Q80897.png)
$$\bigtriangleup'=\left[\sqrt{\mathrm{2}}.\mathrm{cos}\alpha\right]^{\mathrm{2}} −\mathrm{2cos2}\alpha= \\ $$$$\:\:\:\:\:\:=\mathrm{2cos}^{\mathrm{2}} \alpha−\mathrm{2}\left(\mathrm{cos}^{\mathrm{2}} \alpha−\mathrm{sin}^{\mathrm{2}} \alpha\right)=\mathrm{2sin}^{\mathrm{2}} \alpha \\ $$$$\Rightarrow\boldsymbol{\mathrm{x}}=\frac{−\boldsymbol{\mathrm{b}}'\pm\sqrt{\bigtriangleup}}{\boldsymbol{\mathrm{a}}}=\frac{\sqrt{\mathrm{2}}.\boldsymbol{\mathrm{cos}\alpha}\pm\sqrt{\mathrm{2}}\boldsymbol{\mathrm{sin}\alpha}}{\mathrm{2}}= \\ $$$$\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}\alpha}\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\boldsymbol{\mathrm{sin}\alpha}=\boldsymbol{\mathrm{cos}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\pm\boldsymbol{\alpha}\right)\:\:\:. \\ $$