Question

Por Favorrr me ajudem!!
Resolva em R as equações abaixo

a) 2 cos (x) - raiz quadrada de 3 =0

b) 4 sen (t) - raiz quadrada de 3 = 2 sen (t)

c)
$\frac{ \cos( \alpha ) }{ \sqrt{2} } + \frac{1}{2} = 0$
d)
$\frac{1}{3} \tan(x) = \frac{1}{ \sqrt{3} }$

Answer 1

Resposta:

$a) \ 2cos(x)-\sqrt3=0\\2cos(x)=\sqrt3\\cos(x)=\frac{\sqrt3}{2}\\S=\{x\in R/x=\frac{\pi}{6}+2k\pi \ ou \ x=\frac{11\pi}{6}+2k\pi\}$

$b) \ 4sen(t)-\sqrt3=2sen(t)\\4sen(t)-2sen(t)=\sqrt3\\2sen(t)=\sqrt3\\sen(t)=\frac{\sqrt3}{2}\\S=\{t \in R/t=\frac{\pi}{3}+2k\pi \ ou \ t=\frac{2\pi}{3}+2k\pi\}$

$c) \  \frac{cos(\alpha)}{\sqrt2}+\frac{1}{2}=0\\\\\frac{cos(\alpha)}{\sqrt2}=-\frac{1}{2}\\cos(\alpha)=-\frac{\sqrt2}{2}\\S=\{x \in R/\alpha=\frac{3\pi}{2}+2k\pi \ ou \ \alpha=\frac{5\pi}{2}+2k\pi\}$

$d) \ \frac{1}{3}tan(x)=\frac{1}{\sqrt3}\\tg(x)=\frac{3}{\sqrt3}=\sqrt3\\S=\{x \in R/  x=\frac{\pi}{3}+2k\pi \ ou \ x=\frac{4\pi}{3}+2k\pi\}$

Answer 2

Explicação passo-a-passo:

a)

$\sf 2cos~(x)-\sqrt{3}=0$

$\sf 2cos~(x)=\sqrt{3}$

$\sf cos~(x)=\dfrac{\sqrt{3}}{2}$

• $\sf x=30^{\circ}~\Rightarrow~x=\dfrac{\pi}{6}~rad$

• $\sf x=330^{\circ}~\Rightarrow~x=\dfrac{11\pi}{6}~rad$

b)

$\sf 4sen~(t)-\sqrt{3}=2sen(t)$

$\sf 4sen~(t)-2sen~(t)=\sqrt{3}$

$\sf 2sen~(t)=\sqrt{3}$

$\sf sen~(t)=\dfrac{\sqrt{3}}{2}$

• $\sf x=60^{\circ}~\Rightarrow~x=\dfrac{\pi}{3}~rad$

• $\sf x=120^{\circ}~\Rightarrow~x=\dfrac{2\pi}{3}~rad$

c)

$\sf \dfrac{cos~(\alpha)}{\sqrt{2}}+\dfrac{1}{2}=0$

$\sf cos~(\alpha)+\dfrac{\sqrt{2}}{2}=0$

$\sf cos~(\alpha)=\dfrac{-\sqrt{2}}{2}$

• $\sf \alpha=135^{\circ}~\Rightarrow~\alpha=\dfrac{3\pi}{4}~rad$

• $\sf \alpha=225^{\circ}~\Rightarrow~\alpha=\dfrac{5\pi}{4}~rad$

d)

$\sf \dfrac{tg~(x)}{3}=\dfrac{1}{\sqrt{3}}$

$\sf \dfrac{tg~(x)}{3}=\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}$

$\sf \dfrac{tg~(x)}{3}=\dfrac{\sqrt{3}}{3}$

$\sf \dfrac{tg~(x)}{3}=3\cdot\dfrac{\sqrt{3}}{3}$

$\sf \dfrac{tg~(x)}{3}=\sqrt{3}$

• $\sf x=60^{\circ}~\Rightarrow~x=\dfrac{\pi}{3}~rad$

• $\sf x=240^{\circ}~\Rightarrow~x=\dfrac{4\pi}{3}~rad$