👤

Salut, ma puteti ajuta la problema numarul 621...?



Salut Ma Puteti Ajuta La Problema Numarul 621 class=

Răspuns :

[tex]2\arcsin x = \arccos(2x) \\ \\ \arccos(2x) \geq 0 \Rightarrow 2\arcsin x\geq 0 \Rightarrow x\geq 0\\ \\\\ \\2\arcsin x = \arccos(2x)\Big|\sin(~)\\ \\ \sin(2\arcsin x) = \sin(\arccos(2x)) \\ \\ 2\sin(\arcsin x)\cos (\arcsin x) = \sqrt{1-(2x)^2}\\ 2x\sqrt{1-x^2} = \sqrt{1-(2x)^2}\\ 4x^2(1-x^2) = 1-(2x)^2 \\ 4x^2(1-x^2)= 1-4x^2 \\ \\ x \in [0,1] \Rightarrow x^2 \in [0,1] \Rightarrow t\in [0,1]\\ \\ 4t(1-t) = 1-4t \\ 4t - 4t^2 = 1-4t \\ 4t^2-8t+1 = 0 \\ (2t-2)^2-3 = 0 \\ 4(t-1)^2 = 3\\[/tex]

[tex](t-1)^2= \dfrac{3}{4}\\ \\ (1)~~t-1 = \dfrac{\sqrt 3}{2} \Rightarrow t = \dfrac{\sqrt 3}{2}+1,~~~ t > 1 \quad (F) \\ (2)~~t-1 = -\dfrac{\sqrt 3}{2} \Rightarrow t = \dfrac{2-\sqrt 3}{2}\\ \\ \Rightarrow x = +\sqrt{\dfrac{2-\sqrt 3}{2}} = \sqrt{\Big(\dfrac{\sqrt 3 - 1}{2}\Big)^2} = \Big|\dfrac{\sqrt 3 - 1}{2}\Big| = \dfrac{\sqrt 3-1}{2}[/tex]

Raspuns corect (E)