Question

Calculați :
1 + 3cos x = cos 2x

Answer 1

Răspuns:

$x \in \Big\{2k\pi + \frac{2\pi}{3}, 2k\pi - \frac{2\pi}{3}\Big|k\in \mathbb{Z}\Big\}$

Explicație pas cu pas:

$1 + 3\cos x = \cos {2x}\\ \\ 1+3\cos x = 2cos^2 \:x - 1\\ \\ \textrm{Fie }y = \cos x\\ \\ 1 + 3y = 2y^2 - 1\\ \\ 2y^2 - 3y - 2 = 0\\ \\ \Delta = 9 + 4\cdot 4 = 9+16 = 25\\ \\ \sqrt{\Delta} = 5\\ \\ y_{1,2} = \frac{3\pm 5}{4} \\ \\ y_1 = \frac{-2}{4} = -\frac{1}{2}\\ \\ y_2 = \frac{8}{4} = 2\\ \\ \cos x \in [-1,1]\implies y \in [-1; 1] \implies y = -\frac{1}{2}\\ \\ \implies \cos x = -\frac{1}{2} \implies x \in \Big\{2k\pi + \frac{2\pi}{3}, 2k\pi - \frac{2\pi}{3}\Big|k\in \mathbb{Z}\Big\}$

Answer 2

cos2x=2cos²x-1

1+3cosx=2cos²x-1

2cos²x-3cosx-2=0

Notam cosx=t

2t^2-3t-2=0

Δ=(-3)²-4·2·(-2)=9+16=25 ⇒√Δ=5

$t_{1,2}=\frac{-(-3)\pm 5}{2\cdot2}\\ \\t_{1}=\frac{-(-3)+5}{2\cdot2}=\frac{8}{4}=2>1, \ NU \ CONVINE \ t=cosx\in [-1;1]\\ \\t_{2}=\frac{-(-3)-5}{2\cdot2}=\frac{-2}{4}=\frac{-1}{2}\\ \\ cos x=\frac{-1}{2}$