Question Number 208945 by hardmath last updated on 28/Jun/24
![4 cos^2 x − 4 cos^2 3x cos x + cos^2 3x = 0 [ 0 ; (π/2) ] Find: x = ?](https://www.tinkutara.com/question/Q208945.png)
$$\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{x}\:−\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{3x}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{3x}\:=\:\mathrm{0} \\ $$$$\left[\:\mathrm{0}\:;\:\frac{\pi}{\mathrm{2}}\:\right] \\ $$$$\mathrm{Find}:\:\:\mathrm{x}\:=\:? \\ $$
Answered by Berbere last updated on 28/Jun/24
![4cos^2 (x)+cos^2 (3x)≥4∣cos(3x)cos(x)∣≥4∣cos^2 (3x)cos(x)∣ 4cos^2 (x)+cos^2 (3x)−4cos^2 (3x)cos(x)=0 ⇒4∣cos(3x)cos(x)∣−4cos^2 (3x)cos(x)=0 ⇒∣cos(3x)cos(x)∣=cos^2 (3x)cos(x) ⇒∣cos(3x)∣cos(x)−cos^2 (3x)cos(x)=0 cos(x)(∣cos(3x)∣−cos^2 (3x))=0 x=(π/2),or ∣cos(3x)∣=cos(3x).cos(3x) ⇒cos(3x)∈{−1,1} cos(3x)⇒3x=kπ x=((kπ)/3);⇒x=0,(π/3) after tchek]x∈{(π/3),(π/2)}](https://www.tinkutara.com/question/Q208950.png)
$$\mathrm{4}{cos}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\geqslant\mathrm{4}\mid{cos}\left(\mathrm{3}{x}\right){cos}\left({x}\right)\mid\geqslant\mathrm{4}\mid{cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right){cos}\left({x}\right)\mid \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right)−\mathrm{4}{cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right){cos}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\mid{cos}\left(\mathrm{3}{x}\right){cos}\left({x}\right)\mid−\mathrm{4}{cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right){cos}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow\mid{cos}\left(\mathrm{3}{x}\right){cos}\left({x}\right)\mid={cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right){cos}\left({x}\right) \\ $$$$\Rightarrow\mid{cos}\left(\mathrm{3}{x}\right)\mid{cos}\left({x}\right)−{cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right){cos}\left({x}\right)=\mathrm{0} \\ $$$${cos}\left({x}\right)\left(\mid{cos}\left(\mathrm{3}{x}\right)\mid−{cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\right)=\mathrm{0} \\ $$$${x}=\frac{\pi}{\mathrm{2}},{or}\:\mid{cos}\left(\mathrm{3}{x}\right)\mid={cos}\left(\mathrm{3}{x}\right).{cos}\left(\mathrm{3}{x}\right) \\ $$$$\Rightarrow{cos}\left(\mathrm{3}{x}\right)\in\left\{−\mathrm{1},\mathrm{1}\right\} \\ $$$$\:{cos}\left(\mathrm{3}{x}\right)\Rightarrow\mathrm{3}{x}={k}\pi \\ $$$${x}=\frac{{k}\pi}{\mathrm{3}};\Rightarrow{x}=\mathrm{0},\frac{\pi}{\mathrm{3}} \\ $$$$\left.{after}\:{tchek}\right]{x}\in\left\{\frac{\pi}{\mathrm{3}},\frac{\pi}{\mathrm{2}}\right\} \\ $$
Commented by hardmath last updated on 28/Jun/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Commented by Berbere last updated on 28/Jun/24
$${Withe}\:{Pleasur}\:{God}\:{bless}\:{You} \\ $$