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Question Number 207561 by hardmath last updated on 18/May/24

4 sin^2 x + sin 2x = 2 find: x = ?

$$\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{sin}\:\mathrm{2}\boldsymbol{\mathrm{x}}\:\:=\:\:\mathrm{2} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Answered by mathzup last updated on 18/May/24

e⇔4((1−cos(2x))/2) +sin(2x)=2 ⇔ 2−2cos(2x)+sin(2x)=2 ⇔ sin(2x)−2cos(2x)=0 ⇔ (√5){−(2/( (√5)))cos(2x)+(1/( (√5)))sin(2x)}=0 let cosθ=−(2/( (√5))) et sinθ=(1/( (√5))) ⇒ tanθ=−(1/2) ⇒θ=−arctan((1/2)) e ⇔cos(2x)cosθ+sin(2x)sinθ=0 ⇔cos(2x−θ)=0 =cos((π/2)+kπ) ⇔2x−θ=(π/2)+kπ ou 2x−θ=−(π/2)+kπ ⇔2x=(π/2)+θ+kπ ou 2x=θ−(π/2)+kπ ⇔x=(π/4)−(1/2)arctan((1/2))+((kπ)/2)ou x=−(π/4)+(1/2)arctan((1/2))+((kπ)/2) k∈Z

$${e}\Leftrightarrow\mathrm{4}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\:+{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}\:\Leftrightarrow \\ $$$$\mathrm{2}−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)+{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}\:\Leftrightarrow \\ $$$${sin}\left(\mathrm{2}{x}\right)−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)=\mathrm{0}\:\Leftrightarrow \\ $$$$\sqrt{\mathrm{5}}\left\{−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}{cos}\left(\mathrm{2}{x}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{sin}\left(\mathrm{2}{x}\right)\right\}=\mathrm{0} \\ $$$${let}\:{cos}\theta=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:{et}\:{sin}\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$${tan}\theta=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\theta=−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${e}\:\Leftrightarrow{cos}\left(\mathrm{2}{x}\right){cos}\theta+{sin}\left(\mathrm{2}{x}\right){sin}\theta=\mathrm{0} \\ $$$$\Leftrightarrow{cos}\left(\mathrm{2}{x}−\theta\right)=\mathrm{0}\:={cos}\left(\frac{\pi}{\mathrm{2}}+{k}\pi\right) \\ $$$$\Leftrightarrow\mathrm{2}{x}−\theta=\frac{\pi}{\mathrm{2}}+{k}\pi\:{ou}\:\mathrm{2}{x}−\theta=−\frac{\pi}{\mathrm{2}}+{k}\pi \\ $$$$\Leftrightarrow\mathrm{2}{x}=\frac{\pi}{\mathrm{2}}+\theta+{k}\pi\:{ou}\:\mathrm{2}{x}=\theta−\frac{\pi}{\mathrm{2}}+{k}\pi \\ $$$$\Leftrightarrow{x}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{{k}\pi}{\mathrm{2}}{ou} \\ $$$${x}=−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{{k}\pi}{\mathrm{2}} \\ $$$${k}\in{Z} \\ $$

Commented by hardmath last updated on 19/May/24

thank you very much

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$

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