Question

Am nevoie de ajutor la astea 2 exercitii​

Answer 1

$4x^4+2x^2-6=0\implies4\left(x^2\right)^2+2x^2-6=0$

Notam $x^2=t$

Atunci avem: $4t^2+2t-6=0$

$\Delta=2^2-4\times4\times(-6)=4+96=100$

$t_1=\cfrac{-2+\sqrt{100}}{2\times4}=\cfrac{-2+10}{8}=1$

$\implies x^2=1\implies \boxed{x=\pm1}$

$t_2=\cfrac{-2-\sqrt{100}}{2\times4}=\cfrac{-12}{8}=-\cfrac{3}{2}$

$\implies x^2=-\cfrac{3}{2}$, dar $x^2 > 0$, deci aici nu avem solutii

Inseamna ca singurele solutii sunt $1$, respectiv $-1$.



$\cfrac{sin(x)}{cos(x)}=1\implies sin(x)=cos(x)$

Dar $x\in(0, \ \cfrac{\pi}{2})$, deci singura solutie este $\boxed{x=\cfrac{\pi}{4}}$

Answer 2

Răspuns:

Explicație pas cu pas: