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Question Number 167486 by greogoury55 last updated on 18/Mar/22

cos^2 x = cos (((4x)/3)) ; 0≤x≤2π

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} {x}\:=\:\mathrm{cos}\:\left(\frac{\mathrm{4}{x}}{\mathrm{3}}\right)\:;\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi \\ $$

Answered by mr W last updated on 18/Mar/22

cos 2x+1=2 cos ((4x)/3) cos ((6x)/3)+1=2 cos ((4x)/3) let t=((2x)/3) cos 3t+1=2 cos 2t 4 cos^3 t−3 cos t+1=4 cos^2 t−2 (4 cos^2 t−3)(cos t−1)=0 ⇒cos t=1 ⇒t=2kπ ⇒cos t=±((√3)/2) ⇒t=kπ±(π/6) ⇒x=3kπ ⇒x=((3kπ)/2)±(π/4)

$$\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{4}{x}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{6}{x}}{\mathrm{3}}+\mathrm{1}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{4}{x}}{\mathrm{3}} \\ $$$${let}\:{t}=\frac{\mathrm{2}{x}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\mathrm{3}{t}+\mathrm{1}=\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{t} \\ $$$$\mathrm{4}\:\mathrm{cos}^{\mathrm{3}} \:{t}−\mathrm{3}\:\mathrm{cos}\:{t}+\mathrm{1}=\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{t}−\mathrm{2} \\ $$$$\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{t}−\mathrm{3}\right)\left(\mathrm{cos}\:{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:{t}=\mathrm{1}\:\Rightarrow{t}=\mathrm{2}{k}\pi \\ $$$$\Rightarrow\mathrm{cos}\:{t}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow{t}={k}\pi\pm\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow{x}=\mathrm{3}{k}\pi \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}{k}\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{4}} \\ $$

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