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Question Number 121303 by talminator2856791 last updated on 06/Nov/20
cos x − (√3) sin x = 2 cos 2x
$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:{x}\:−\:\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\: \\ $$$$\: \\ $$
Commented by liberty last updated on 06/Nov/20
⇒cos x−(√3) sin x = k cos (x−α) k=(√(1^2 +(−(√3))^2 )) = 2 tan α=((−(√3))/1) (4^(th) quadrant) α = 300° ⇒2cos (x−300°) = 2cos 2x ⇒cos 2x= cos (x−300°)
$$\Rightarrow\mathrm{cos}\:\mathrm{x}−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{k}\:\mathrm{cos}\:\left(\mathrm{x}−\alpha\right) \\ $$$$\:\mathrm{k}=\sqrt{\mathrm{1}^{\mathrm{2}} +\left(−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$$$\:\mathrm{tan}\:\alpha=\frac{−\sqrt{\mathrm{3}}}{\mathrm{1}}\:\left(\mathrm{4}^{\mathrm{th}} \:\mathrm{quadrant}\right)\: \\ $$$$\:\alpha\:=\:\mathrm{300}°\: \\ $$$$\Rightarrow\mathrm{2cos}\:\left(\mathrm{x}−\mathrm{300}°\right)\:=\:\mathrm{2cos}\:\mathrm{2x}\: \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2x}=\:\mathrm{cos}\:\left(\mathrm{x}−\mathrm{300}°\right)\: \\ $$$$ \\ $$
Answered by Bird last updated on 06/Nov/20
cosx−(√3)sinx=2cos(2x)⇒ 2((1/2)cosx−((√3)/2)sinx)=2cos(2x) ⇒cosxcos((π/3))−sinx sin((π/3))=cos(2x) ⇒cos(x+(π/3))=cos(2x) ⇒ x+(π/3)=2x+2kπ or x+(π/3)=−2x+2kπ ⇒−x=−(π/3)+2kπ or 3x=−(π/3)+2kπ ⇒ x=(π/3)−2kπ or x=−(π/9)+((2kπ)/3) k∈Z
$${cosx}−\sqrt{\mathrm{3}}{sinx}=\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\Rightarrow \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}{cosx}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sinx}\right)=\mathrm{2}{cos}\left(\mathrm{2}{x}\right) \\ $$$$\Rightarrow{cosxcos}\left(\frac{\pi}{\mathrm{3}}\right)−{sinx}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)={cos}\left(\mathrm{2}{x}\right) \\ $$$$\Rightarrow{cos}\left({x}+\frac{\pi}{\mathrm{3}}\right)={cos}\left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$${x}+\frac{\pi}{\mathrm{3}}=\mathrm{2}{x}+\mathrm{2}{k}\pi\:{or}\:{x}+\frac{\pi}{\mathrm{3}}=−\mathrm{2}{x}+\mathrm{2}{k}\pi \\ $$$$\Rightarrow−{x}=−\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\:{or}\:\mathrm{3}{x}=−\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\:\Rightarrow \\ $$$${x}=\frac{\pi}{\mathrm{3}}−\mathrm{2}{k}\pi\:{or}\:{x}=−\frac{\pi}{\mathrm{9}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}} \\ $$$${k}\in{Z} \\ $$$$ \\ $$

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