Question

Os valores reais de m que satisfazem \sf \dpi{90} m-\sqrt3\cdot cotg\,x=-1, com \sf \dpi{90} x\in \left]\frac{\pi}{6},\frac{\pi}{3}\right[, pertencem ao intervalo ]0, 1[. ]0, 2[. ]0, π[. ]0, 2π[.

Answer 1

Resposta:

$tgx=m-2 \\ cotgx= \frac{m}{3} ==\ \textgreater \ \frac{1}{tgx} = \frac{m}{3} ==\ \textgreater \ 3=m*tgx==\ \textgreater \ \\ 3=m(m-2) ==\ \textgreater \ m^2-2m-3=0 \\ m_{12}= \frac{-(-2)+/- \sqrt{(-2)^2-4(1)(-3)} }{2} = \frac{2+/-4}{2} \\ m_{1} =3 \\ m_{2} =-1$

Answer 2

Resposta:

$m\in\,\,]0, 2[$

Explicação passo a passo:

$m-\sqrt3\cdot cotg\,x=-1$

$cotg\,x=\frac{m+1}{\sqrt3}\Rightarrow tg\, x=\frac{\sqrt3}{m+1}$

$tg\, x \in \left]\frac{\sqrt3}{3}, \sqrt3\right[\Rightarrow\frac{\sqrt3}{3} < tg\,x < \sqrt3\Rightarrow\frac{\sqrt3}{3} < \frac{\sqrt3}{m+1} < \sqrt3$

$\frac{1}{3} < \frac{1}{m+1} < 1\Rightarrow m+1 < 3 < 3m+3\Rightarrow0 < m < 2$

$m\in\,\,]0, 2[$

Espero ter ajudado! ∩ω∩