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Question Number 87488 by jagoll last updated on 04/Apr/20
sin^4 x + sin^4 (x+(π/4)) = (1/4) x ∈ [ 0,2π ]
$$\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}\:+\:\mathrm{sin}\:^{\mathrm{4}} \left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{x}\:\in\:\left[\:\mathrm{0},\mathrm{2}\pi\:\right]\: \\ $$
Commented by john santu last updated on 05/Apr/20
(2sin^2 x)^2 +(2sin^2 (x+(π/4)))^2 = 1 (1−cos 2x)^2 + (1−cos (2x+(π/2)))^2 =1 (1−cos 2x)^2 + (1+sin 2x)^2 = 1 1−2cos 2x +cos^2 2x+1+2sin 2x +sin^2 2x = 1 2sin 2x−2cos 2x = −2 cos 2x−sin 2x = 1 cos (2x+(π/4)) = (1/( (√2))) = cos (π/4) ⇒2x = −(π/4) ± (π/4)+2πn , n ∈Z ⇒ x = −(π/8) ± (π/8) + πn (+) ⇒ x = 0, π ,2π (−)⇒x = −(π/4)+πn ; x = ((3π)/4), ((7π)/4)
$$\left(\mathrm{2sin}^{\mathrm{2}} \:\mathrm{x}\right)^{\mathrm{2}} +\left(\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\right)^{\mathrm{2}} =\:\mathrm{1}\: \\ $$$$\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{2}} +\:\left(\mathrm{1}−\mathrm{cos}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{2}}\right)\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{2}} \:+\:\left(\mathrm{1}+\mathrm{sin}\:\mathrm{2x}\right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{2cos}\:\mathrm{2x}\:+\mathrm{cos}\:^{\mathrm{2}} \:\mathrm{2x}+\mathrm{1}+\mathrm{2sin}\:\mathrm{2x}\:+\mathrm{sin}\:^{\mathrm{2}} \:\mathrm{2x}\:=\:\mathrm{1} \\ $$$$\mathrm{2sin}\:\mathrm{2x}−\mathrm{2cos}\:\mathrm{2x}\:=\:−\mathrm{2} \\ $$$$\mathrm{cos}\:\mathrm{2x}−\mathrm{sin}\:\mathrm{2x}\:=\:\mathrm{1}\: \\ $$$$\mathrm{cos}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{2x}\:=\:−\frac{\pi}{\mathrm{4}}\:\pm\:\frac{\pi}{\mathrm{4}}+\mathrm{2}\pi\mathrm{n}\:,\:\mathrm{n}\:\in\mathbb{Z} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:−\frac{\pi}{\mathrm{8}}\:\pm\:\frac{\pi}{\mathrm{8}}\:+\:\pi\mathrm{n}\: \\ $$$$\left(+\right)\:\Rightarrow\:\mathrm{x}\:=\:\mathrm{0},\:\pi\:,\mathrm{2}\pi \\ $$$$\left(−\right)\Rightarrow\mathrm{x}\:=\:−\frac{\pi}{\mathrm{4}}+\pi\mathrm{n}\:;\:\mathrm{x}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}},\:\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$$$ \\ $$
Answered by mind is power last updated on 04/Apr/20
sin(x+(π/4))=((√2)/2)(sin(x)+cos(x)) sin^2 (x+(π/4))=(1/2)(1+sin(2x)) sin^4 (x+(π/4))=(1/4)(1+sin^2 (2x)+2sin(2x)) ⇔sin^4 (x)+((2sin(2x)+sin^2 (2x))/4)=0 ⇔sin(x)(4sin^3 (x)+4cos(x)+4sin(x)cos^2 (x))=0 ⇔4sin(x)(sin(x)+cos(x))=0 ⇒4(√2)sin(x)sin(x+(π/4))=0 x∈{0,π,2π,((3π)/4),((7π)/4)}
$${sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({sin}\left({x}\right)+{cos}\left({x}\right)\right) \\ $$$${sin}^{\mathrm{2}} \left({x}+\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{sin}\left(\mathrm{2}{x}\right)\right) \\ $$$${sin}^{\mathrm{4}} \left({x}+\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)+\mathrm{2}{sin}\left(\mathrm{2}{x}\right)\right) \\ $$$$\Leftrightarrow{sin}^{\mathrm{4}} \left({x}\right)+\frac{\mathrm{2}{sin}\left(\mathrm{2}{x}\right)+{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{4}}=\mathrm{0} \\ $$$$\Leftrightarrow{sin}\left({x}\right)\left(\mathrm{4}{sin}^{\mathrm{3}} \left({x}\right)+\mathrm{4}{cos}\left({x}\right)+\mathrm{4}{sin}\left({x}\right){cos}^{\mathrm{2}} \left({x}\right)\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{4}{sin}\left({x}\right)\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\sqrt{\mathrm{2}}{sin}\left({x}\right){sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{0} \\ $$$${x}\in\left\{\mathrm{0},\pi,\mathrm{2}\pi,\frac{\mathrm{3}\pi}{\mathrm{4}},\frac{\mathrm{7}\pi}{\mathrm{4}}\right\} \\ $$
Answered by TANMAY PANACEA. last updated on 04/Apr/20
(((1−cos2x)/2))^2 +(((1−cos((π/2)+2x))/2))^2 =(1/4) ((1−2cos2x+cos^2 2x)/4)+(((1+sin2x)/2))^2 =(1/4) 1−2cos2x+cos^2 2x+1+2sin2x+sin^2 2x=1 3−2cos2x+2sin2x=1 1−cos2x+sin2x=0 1=(cos2x−sin2x)^2 1=1−sin4x sin4x=0=sin0→x=0
$$\left(\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}−\mathrm{2}{cos}\mathrm{2}{x}+{cos}^{\mathrm{2}} \mathrm{2}{x}}{\mathrm{4}}+\left(\frac{\mathrm{1}+{sin}\mathrm{2}{x}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{1}−\mathrm{2}{cos}\mathrm{2}{x}+{cos}^{\mathrm{2}} \mathrm{2}{x}+\mathrm{1}+\mathrm{2}{sin}\mathrm{2}{x}+{sin}^{\mathrm{2}} \mathrm{2}{x}=\mathrm{1} \\ $$$$\mathrm{3}−\mathrm{2}{cos}\mathrm{2}{x}+\mathrm{2}{sin}\mathrm{2}{x}=\mathrm{1} \\ $$$$\mathrm{1}−{cos}\mathrm{2}{x}+{sin}\mathrm{2}{x}=\mathrm{0} \\ $$$$\mathrm{1}=\left({cos}\mathrm{2}{x}−{sin}\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}=\mathrm{1}−{sin}\mathrm{4}{x} \\ $$$${sin}\mathrm{4}{x}=\mathrm{0}={sin}\mathrm{0}\rightarrow{x}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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