Question

Sa se rezolve ecuatia de mai jos

2·cos(2x)+cos(6x)=0

Answer 1

$\cos(3x) = 4\cos^3(x) - 3\cos x \\ \\\\ 2\cos(2x)+\cos\Big[3\cdot (2x)\Big]=0\\ \\ 2\cos(2x)+4\cos^3(2x)-3\cos(2x) = 0 \\ \\ \cos(2x) = t,\quad t\in [-1,1]\\ \\2t+4t^3-3t = 0 \\ 4t^3-t =0\\ t(4t^2-1) = 0 \\ \\ (1)\quad t = 0 \Rightarrow \cos(2x) = 0 \Rightarrow \cos(2x) = \cos(\frac{\pi}{2})\Rightarrow \\ \Rightarrow 2x = 2k\pi\pm \dfrac{\pi}{2}\Rightarrow x = k\pi\pm \dfrac{\pi}{4}$

$(2)\quad 4t^2-1 = 0 \Rightarrow 4t^2 = 1 \Rightarrow t = \pm \dfrac{1}{2}\Rightarrow \\ \Rightarrow \cos(2x) = \pm \dfrac{1}{2} \Rightarrow 2x =2k\pi\pm \arccos\Big(\pm\dfrac{1}{2}\Big) \Rightarrow \\ \\ \Rightarrow 2x = 2k\pi\pm \dfrac{\pi}{3}\quad sau\quad 2x = 2k\pi\pm\dfrac{2\pi}{3}\\ \Rightarrow x = k\pi\pm \dfrac{\pi}{6}\quad sau \quad x = k\pi \pm \dfrac{\pi}{3} \\ \\ \\\Rightarrow \boxed{x = \Big\{k\pi\pm \dfrac{\pi}{6},~k\pi\pm \dfrac{\pi}{4},~k\pi\pm \dfrac{\pi}{3}\Big\},\quad k\in \mathbb{Z}}$

Answer 2

Răspuns:

Explicație pas cu pas:

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