Question

No intervalo de [0, 2pi] resolva a inequação:


tg²(x) > tg(x)

Answer 1

$\large\boxed{\begin{array}{l}\sf\\\sf tg^2(x)>tg(x)\\\sf fac_{\!\!,}a\,u=tg(x)\\\sf u^2>u\\\sf u^2-u>0\\\underline{\rm fac_{\!\!,}a}\\\sf f(u)=u^2-u\\\underline{\rm zeros\,de\,f(u):}\\\sf u^2-u=0\\\sf u\cdot(u-1)=0\\\sf u=0\\\sf u-1=0\\\sf u=1\\\sf f(u)>0\iff u<0~~ou~~u>1\end{array}}$

$\large\boxed{\begin{array}{l}\sf tg(x)<0\\\sf \dfrac{3\pi}{2}<x<0.\\\sf tg(x)>1\\\sf\dfrac{\pi}{4}<x<\dfrac{\pi}{2}~~ou~~\dfrac{5\pi}{4}<x<\dfrac{3\pi}{2}\\\\\underline{\rm fazendo\,a\,intersecc_{\!\!,}\tilde ao\,temos}\\\\\sf\dfrac{\pi}{4}<x<\dfrac{\pi}{2}~~ou~~\dfrac{3\pi}{2}<x<\pi\end{array}}$