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Question Number 140177 by liberty last updated on 05/May/21
solution set equation sin^6 x + cos^6 x = (1/4)sin^2 2x
$$\:\:\:\mathrm{solution}\:\mathrm{set}\:\mathrm{equation} \\ $$$$\:\:\:\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}\:+\:\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$
Answered by som(math1967) last updated on 05/May/21
(sin^2 x+cos^2 x)^3 −3sin^2 xcos^2 x(sin^2 x+cos^2 x)=(1/4)sin^2 2x 1−(3/4)(2sinxcosx)^2 =(1/4)sin^2 2x 1=(3/4)sin^2 2x+(1/4)sin^2 2x ⇒sin^2 2x=1 sin2x=±1 when sin2x=1 x=(4n+1)(π/4) sin2x=−1 x=(4n−1)(π/4)
$$\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{3}} \\ $$$$\:\:−\mathrm{3}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)=\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{2}{sinxcosx}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\mathrm{1}=\frac{\mathrm{3}}{\mathrm{4}}{sin}^{\mathrm{2}} \mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$\Rightarrow{sin}^{\mathrm{2}} \mathrm{2}{x}=\mathrm{1} \\ $$$${sin}\mathrm{2}{x}=\pm\mathrm{1} \\ $$$${when}\:{sin}\mathrm{2}{x}=\mathrm{1} \\ $$$$\:\:{x}=\left(\mathrm{4}{n}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}} \\ $$$${sin}\mathrm{2}{x}=−\mathrm{1} \\ $$$$\:{x}=\left(\mathrm{4}{n}−\mathrm{1}\right)\frac{\pi}{\mathrm{4}} \\ $$
Answered by liberty last updated on 05/May/21
⇒sin^6 x+cos^6 x=(1/4).4sin^2 x cos^2 x ⇒(sin^2 x+cos^2 x)(sin^4 x−sin^2 xcos^2 x+cos^4 x)=(sin x cos x)^2 ⇒1−3sin^2 x cos^2 x = sin^2 x cos^2 x ⇒4sin^2 x cos^2 x = 1 ⇒(sin 2x+1)(sin 2x−1)=0 ⇒ { ((sin 2x=−1)),((sin 2x=1)) :} ⇒ { ((x=((3π)/4)+nπ)),((x=(π/4)+nπ)) :}
$$\Rightarrow\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{4}}.\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\Rightarrow\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{xcos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}\right)=\left(\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{3sin}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\Rightarrow\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{sin}\:\mathrm{2x}+\mathrm{1}\right)\left(\mathrm{sin}\:\mathrm{2x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{sin}\:\mathrm{2x}=−\mathrm{1}}\\{\mathrm{sin}\:\mathrm{2x}=\mathrm{1}}\end{cases}\:\Rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{3}\pi}{\mathrm{4}}+\mathrm{n}\pi}\\{\mathrm{x}=\frac{\pi}{\mathrm{4}}+\mathrm{n}\pi}\end{cases} \\ $$

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