Question

Resolver a equação 2sen²x+cos x−1=0:
a) para 0 ≤ x < 2pi
b) Em R

se possivel mostre como resolveu quero entender o processo, obrigado

Answer 1

$\mathrm{2\sin^2{x}+\cos{x}-1=0}\\ \mathrm{2(1-\cos^2{x})+\cos{x}-1=0}\\ \mathrm{2-2\cos^2{x}+\cos{x}-1=0}\\ \mathrm{1-2\cos^2{x}+\cos{x}=0}\\ \mathrm{-2\cos^2{x}+\cos{x}+1=0\ \to\ a=-2\ \| \ b=1\ \| \ c=1}\\\\ \mathrm{\cos{x}=\cfrac{-b\pm\sqrt{\Delta}}{2a}\ \to\ \Delta=b^2-4ac}\\\\ \mathrm{\Delta=1^2-4.(-2).1=1+8=9}\\\\ \mathrm{\cos'{x}=\cfrac{-1+\sqrt{9}}{2.(-2)}=\cfrac{-1+3}{-4}=\cfrac{2}{-4}=-\cfrac{1}{2}}\\\\ \mathrm{\cos''{x}=\cfrac{-1-\sqrt{9}}{2.(-2)}=\cfrac{-1-3}{-4}=\cfrac{-4}{-4}=1}$

$\mathrm{\cos{x}=\{\cos'{x},\cos''{x}\}=\{-\dfrac{1}{2},1\}}\\\\\\ \textbf{a)}\ \mathrm{Solu\c{c}\~oes\ em\ 0\leq x\ \textless \ 2\pi}\\\\ \mathbf{Para\ \cos{x}=-\dfrac{1}{2},\ temos\ 2\ solu\c{c}\~oes:}\\\\ \mathrm{x=\{120\º,240\º\}=\{\dfrac{2\pi}{3},\dfrac{4\pi}{3}\}}\\\\ \mathbf{Para\ \cos{x}=1,\ temos\ 1\ solu\c{c}\~ao:}\\\\ \mathrm{x=\{0\º\}=\{0\}}\\\\ \mathrm{Portanto}\ \to\ \mathbf{x=\{0,\dfrac{2\pi}{3},\dfrac{4\pi}{3}\}}$

$\mathbf{b)}\ \mathrm{Solu\c{c}\~oes\ em}\ \mathbb{R}\\\\ \mathbf{Para\ \cos{x}=-\dfrac{1}{2}:}\\\\ \mathrm{x=\{120\º+2k\pi,240\º+2k\pi\}=\{\dfrac{2\pi}{3}+2k\pi,\dfrac{4\pi}{3}+2k\pi\}}\\\\ \mathbf{Para\ \cos{x}=1:}\\\\ \mathrm{x=\{0\°+2k\pi\}=\{2k\pi\}}\\\\ \mathrm{Portanto}\ \to\ \mathbf{x=\{2k\pi,\dfrac{2\pi}{3}+2k\pi,\dfrac{4\pi}{3}+2k\pi\}}$