Menu Close

sin-4-x-4cos-2-x-cos-4-x-4sin-2-x-1-2-for-x-0-2pi-

Tinku Tara 0 Comments
Pin




Question Number 140378 by benjo_mathlover last updated on 07/May/21
(√(sin^4 x+4cos^2 x))−(√(cos^4 x+4sin^2 x)) = (1/2) for x∈ [ 0,2π ]
$$\sqrt{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}}−\sqrt{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}+\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{for}\:\mathrm{x}\in\:\left[\:\mathrm{0},\mathrm{2}\pi\:\right] \\ $$
Commented by john_santu last updated on 07/May/21
(√(sin^4 x+4cos^2 x)) > (√(cos^4 x+4sin^2 x)) ⇒sin^2 x(sin^2 x−4) > cos^2 x(cos^2 x−4) ⇒sin^2 x >cos^2 x ⇒sin^2 x >1−sin^2 x ((√2) sin x−1)((√2) sin x+1)>0 ⇒ sin x <−(1/( (√2))) ∪ sin x > (1/( (√2)))
$$\sqrt{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{4cos}\:^{\mathrm{2}} {x}}\:>\:\sqrt{\mathrm{cos}\:^{\mathrm{4}} {x}+\mathrm{4sin}\:^{\mathrm{2}} {x}} \\ $$$$\Rightarrow\mathrm{sin}\:^{\mathrm{2}} {x}\left(\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{4}\right)\:>\:\mathrm{cos}\:^{\mathrm{2}} {x}\left(\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{sin}\:^{\mathrm{2}} {x}\:>\mathrm{cos}\:^{\mathrm{2}} {x}\: \\ $$$$\Rightarrow\mathrm{sin}\:^{\mathrm{2}} {x}\:>\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$$\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:{x}+\mathrm{1}\right)>\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:<−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\cup\:\mathrm{sin}\:{x}\:>\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\: \\ $$
Answered by john_santu last updated on 07/May/21
(√(sin^4 x+4−4sin^2 x)) −(√(cos^4 x+4−4cos^2 x)) =(1/2) (√((2−sin^2 x)^2 ))−(√((2−cos^2 x)^2 )) =(1/2) 2−sin^2 x+cos^2 x−2 = (1/2) cos 2x = (1/2) cos 2x = cos (π/3) 2x= { ((−(π/3) +2nπ)),(((π/3)+2nπ)) :}⇔ { ((x=−(π/6)+nπ)),((x=(π/6)+nπ)) :} x = { (π/6) , ((5π)/6) , ((7π)/3) , ((11π)/6) }
$$\sqrt{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{4}−\mathrm{4sin}\:^{\mathrm{2}} {x}}\:−\sqrt{\mathrm{cos}\:^{\mathrm{4}} {x}+\mathrm{4}−\mathrm{4cos}\:^{\mathrm{2}} {x}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sqrt{\left(\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} }−\sqrt{\left(\mathrm{2}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\cancel{\mathrm{2}}−\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}−\cancel{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\:\mathrm{2}{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{3}} \\ $$$$\:\mathrm{2}{x}=\begin{cases}{−\frac{\pi}{\mathrm{3}}\:+\mathrm{2}{n}\pi}\\{\frac{\pi}{\mathrm{3}}+\mathrm{2}{n}\pi}\end{cases}\Leftrightarrow\:\begin{cases}{{x}=−\frac{\pi}{\mathrm{6}}+{n}\pi}\\{{x}=\frac{\pi}{\mathrm{6}}+{n}\pi}\end{cases} \\ $$$${x}\:=\:\left\{\:\frac{\pi}{\mathrm{6}}\:,\:\frac{\mathrm{5}\pi}{\mathrm{6}}\:,\:\frac{\mathrm{7}\pi}{\mathrm{3}}\:,\:\frac{\mathrm{11}\pi}{\mathrm{6}}\:\right\}\: \\ $$
Answered by MJS_new last updated on 07/May/21
(√(sin^4 x +4cos^2 x))−(√(cos^4 x +4sin^2 x))=(1/2) let t=tan x (√(((t/( (√(t^2 +1)))))^4 +4((1/( (√(t^2 +1)))))^2 ))−(√(((1/( (√(t^2 +1)))))^4 +4((t/( (√(t^2 +1)))))^2 ))=(1/2) (√((((t^2 +2)/(t^2 +1)))^2 ))−(√((((2t^2 +1)/(t^2 +1)))^2 ))=(1/2) ((1−t^2 )/(1+t^2 ))=(1/2) ⇒ t=±((√3)/3)=tan x ⇒ x=(π/6)∨x=((5π)/6)∨x=((7π)/6)∨x=((11π)/6)
$$\sqrt{\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{4cos}^{\mathrm{2}} \:{x}}−\sqrt{\mathrm{cos}^{\mathrm{4}} \:{x}\:+\mathrm{4sin}^{\mathrm{2}} \:{x}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{let}\:{t}=\mathrm{tan}\:{x} \\ $$$$\sqrt{\left(\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\right)^{\mathrm{4}} +\mathrm{4}\left(\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\right)^{\mathrm{2}} }−\sqrt{\left(\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\right)^{\mathrm{4}} +\mathrm{4}\left(\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sqrt{\left(\frac{{t}^{\mathrm{2}} +\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }−\sqrt{\left(\frac{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:{t}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\mathrm{tan}\:{x} \\ $$$$\Rightarrow\:{x}=\frac{\pi}{\mathrm{6}}\vee{x}=\frac{\mathrm{5}\pi}{\mathrm{6}}\vee{x}=\frac{\mathrm{7}\pi}{\mathrm{6}}\vee{x}=\frac{\mathrm{11}\pi}{\mathrm{6}} \\ $$
Commented by greg_ed last updated on 07/May/21
right !
$$\mathrm{right}\:! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *