Question

preciso muito de ajuda nisso

se a cossec x= 2√3/3, determine o valor de tg x/2

Answer 1

Resposta:

Explicação passo a passo:

$cossec(x)=\frac{2\sqrt{3} }{3}\\\\\frac{1}{sen(x)} = \frac{2\sqrt{3} }{3}\\\\sen(x)=\frac{3}{2\sqrt{3}}\\\\sen^{2}(x)=(\frac{3 }{2\sqrt{3}})^{2}\\\\sen^{2}(x)=\frac{9}{4*3}\\\\ sen^{2}(x)=\frac{3}{4}\\\\1-cos^{2}(x)= \frac{3}{4}\\\\cos^{2}(x)=1-\frac{3}{4}\\\\cos^{2}(x)=\frac{4}{4} -\frac{3}{4}\\\\cos^{2}(x)=\frac{1}{4} \\\\|cos(x)|=\frac{1}{2}\\\\-------\\\\tg(x)=\frac{sen(x)}{cos(x)}\\\\ |tg(x)|=\frac{\frac{3}{2\sqrt{3}}}{\frac{1}{2} }\\\\|tg(x)|=\frac{3}{\sqrt{3} } \\\\-------$

$tg(2(\frac{x}{2}))=\frac{2tg(\frac{x}{2})}{1-tg^{2}(\frac{x}{2})} \\\\tg(x)=\frac{2tg(\frac{x}{2})}{1-tg^{2}(\frac{x}{2})} \\\\\ tg(x)-tg(x)tg^{2}(\frac{x}{2}) =2tg(\frac{x}{2} )\\\\tg(x)tg^{2}(\frac{x}{2}) +2tg(\frac{x}{2} )-tg(x)=0\\\\tg(\frac{x}{2})=\frac{-2+-\sqrt{4+4tg^{2}(x)} }{2tg(x)}\\\\tg(\frac{x}{2})=\frac{-1+-\sqrt{1+tg^{2}(x)} }{tg(x)}$

$se\ tg(x)> 0, ou\ seja,\ \pi k< x <\frac{\pi }{2}+\pi k,k\ e\ Z\\\\tg(\frac{x}{2})=\frac{-1+-\sqrt{1+(\frac{3}{\sqrt{3} } )^{2}}}{\frac{3}{\sqrt{3}}}\\\\tg(\frac{x}{2})=\frac{-1+-2}{\frac{3}{\sqrt{3}}}\\\\tg(\frac{x}{2})=\frac{\sqrt{3} }{3}\ ou\ tg(\frac{x}{2})=-\sqrt{3}\ (nao\ serve)\\\\se\ tg(x)< 0, ou\ seja,\ \frac{\pi }{2} +\pi k< x< \pi +\pi k,k\ e\ Z\\\\tg(\frac{x}{2})=\frac{-1+-\sqrt{1+(\frac{3}{\sqrt{3} } )^{2}}}{-\frac{3}{\sqrt{3}}}$

$tg(\frac{x}{2})=\frac{-1+-2}{-\frac{3}{\sqrt{3}}}\\\\tg(\frac{x}{2})=-\frac{\sqrt{3} }{3}\ (nao\ serve)\ ou\ tg(\frac{x}{2})=\sqrt{3}\\\\\\Entao\ tg(\frac{x}{2}) = \frac{\sqrt{3} }{3}\ ou\ tg(\frac{x}{2}) = \sqrt{3}$

Answer 2

Resposta:

csc(x) = 2√3/3 = 2/√3

sen(x) = √3/2

x = 60° ou 120°

tg(x/2) = tg(60/2) = tg(30) = √3/3

tg(120/2) = tg(60) = √3