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Resolvamos a seguinte inequação:

[tex]tg\left(x\right) \cdot tg\left(2x\right) \geq 1[/tex]

com [tex]x \in \left[0, 2\pi\right[.[/tex]

Estabeleçamos, inicialmente, seu domínio de validade:

Temos:

[tex]x \in \left[0, 2\pi\right[ \,\,\, \land \,\,\,x \neq \dfrac{\pi}{2} + k\pi, k \in \mathbb{Z}\,\,\, \land \,\,\, x \neq \dfrac{\pi}{4} + \dfrac{k\pi}{2}, k \in \mathbb{Z}[/tex]

[tex]D_v = \left\{x \in \mathbb{R} \, | \, 0 \leq x < 2\pi\right\} \setminus \left\{\dfrac{\pi}{4}, \dfrac{\pi}{2}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{3\pi}{2}, \dfrac{7\pi}{4}\right\}[/tex]

A solução abaixo lança mão da seguinte identidade trigonométrica:

[tex]tg\left(2x\right) = \dfrac{2\cdot tg\left(x\right)}{1 - tg^2\left(x\right)}, \,\forall \, x \in \mathbb{R} \, | \, x \neq \dfrac{\pi}{2}+ k\pi\,\,\,e\,\,\,x \neq \dfrac{\pi}{4} + \dfrac{k\pi}{2}, k \in \mathbb{Z}.[/tex]

Resolução:

[tex]tg\left(x\right) \cdot tg\left(2x\right) \geq 1\\\\\\\Longleftrightarrow tg\left(x\right) \cdot \dfrac{2 \cdot tg\left(x\right)}{1 - tg^2\left(x\right)} \geq 1\\\\\\\Longleftrightarrow \dfrac{2 \cdot tg^2\left(x\right)}{1 - tg^2\left(x\right)} - 1 \geq 0\\\\\\\Longleftrightarrow \dfrac{2 \cdot tg^2\left(x\right) - \left[1 - tg^2\left(x\right)\right]}{1 - tg^2\left(x\right)} \geq 0\\\\\\\Longleftrightarrow \dfrac{3 \cdot tg^2\left(x\right)-1}{1 - tg^2\left(x\right)} \geq 0\,\,\,\,\,\,\,\,\,\,(I)[/tex]

Estudando o sinal da função [tex]f\left[tg\left(x\right)\right] = 3 \cdot tg^2\left(x\right)-1,[/tex] observamos que:

[tex]f = 0 \Longleftrightarrow tg\left(x\right) = \pm \dfrac{\sqrt{3}}{3};\\\\\\ f > 0 \Longleftrightarrow tg\left(x\right) < -\dfrac{\sqrt{3}}{3}\,\,\,\lor\,\,\,tg\left(x\right) > \dfrac{\sqrt{3}}{3};\\\\\\f < 0 \Longleftrightarrow -\dfrac{\sqrt{3}}{3} < tg\left(x\right) < \dfrac{\sqrt{3}}{3}.[/tex]

Estudando o sinal da função [tex]g\left[tg\left(x\right)\right] = 1 - tg^2\left(x\right),[/tex] notamos que:

[tex]g = 0 \Longleftrightarrow tg\left(x\right) = \pm 1;\\\\\\g > 0 \Longleftrightarrow -1 < tg\left(x\right) < 1;\\\\\\g > 0 \Longleftrightarrow tg\left(x\right) < -1 \,\,\, \lor \,\,\, tg\left(x\right) > 1.[/tex]

Para que a desigualdade [tex](I)[/tex] seja verdadeira, devemos observar três possibilidades:

[tex]i) \,\,\, \left(f = 0\right)\,\,\,\,\, \land \,\,\,\,\, \left(g \neq 0\right)\\\\\\\Longleftrightarrow \left(tg\left(x\right) = - \dfrac{\sqrt{3}}{3}\,\,\, \lor \,\,\, tg\left(x\right) = \dfrac{\sqrt{3}}{3} \right) \,\,\,\,\, \land \,\,\,\,\, \left(tg\left(x\right) \neq -1 \,\,\, \land \,\,\, tg\left(x\right) \neq 1 \right)\\\\\\\Longleftrightarrow tg\left(x\right) = - \dfrac{\sqrt{3}}{3}\,\,\, \lor \,\,\, tg\left(x\right) = \dfrac{\sqrt{3}}{3}[/tex]

[tex]ii)\,\,\, \left(f > 0\right) \,\,\,\,\, \land\,\,\,\,\, \left(g > 0 \right)\\\\\\\Longleftrightarrow \left(tg\left(x\right) < -\dfrac{\sqrt{3}}{3}\,\,\,\lor\,\,\,tg\left(x\right) > \dfrac{\sqrt{3}}{3}\right) \,\,\,\,\, \land\,\,\,\,\, \left(-1 < tg\left(x\right) < 1 \right)\\\\\\\Longleftrightarrow -1 < tg\left(x\right) < -\dfrac{\sqrt{3}}{3}\,\,\,\,\, \lor \,\,\,\,\,\dfrac{\sqrt{3}}{3} < tg\left(x\right) < 1[/tex]

[tex]iii)\,\,\, \left(f < 0\right) \,\,\,\,\, \land\,\,\,\,\, \left(g < 0 \right)\\\\\\\Longleftrightarrow \left(-\dfrac{\sqrt{3}}{3} < tg\left(x\right) < \dfrac{\sqrt{3}}{3}\right) \,\,\,\,\, \land\,\,\,\,\, \left(tg\left(x\right) \leq -1 \,\,\, \lor \,\,\,tg\left(x\right) \geq 1\right)\\\\\\\Longleftrightarrow tg\left(x\right) \in \emptyset[/tex]

Estabelecendo a reunião das soluções (i), (ii) e (iii), obtemos:

[tex]-1 < tg\left(x\right) \leq -\dfrac{\sqrt{3}}{3}\,\,\, \lor \,\,\, \dfrac{\sqrt{3}}{3} \leq tg\left(x\right) < 1[/tex]

Resolvamos para [tex]x:[/tex]

[tex]\left(x \in D_v\right) \,\,\,\,\, \land \,\,\,\,\,\left(-1 < tg\left(x\right) \leq -\dfrac{\sqrt{3}}{3}\,\,\, \lor \,\,\, \dfrac{\sqrt{3}}{3} \leq tg\left(x\right) < 1\right)\\\\\\\Longleftrightarrow \left(\dfrac{3\pi}{4} < x \leq \dfrac{5\pi}{6}\,\,\,\lor\,\,\, \dfrac{7\pi}{4} < x \leq \dfrac{11\pi}{6}\right) \,\,\,\,\,\lor\,\,\,\,\,\left(\dfrac{\pi}{6} \leq x < \dfrac{\pi}{4} \,\,\, \lor \,\,\, \dfrac{7\pi}{6} \leq x < \dfrac{5\pi}{4}\right)\\\\\\[/tex]

[tex]\Longleftrightarrow \dfrac{\pi}{6} \leq x < \dfrac{\pi}{4} \,\,\, \lor \,\,\, \dfrac{3\pi}{4} < x \leq \dfrac{5\pi}{6} \,\,\, \lor \,\,\, \dfrac{7\pi}{6} \leq x < \dfrac{5\pi}{4} \,\,\, \lor \,\,\, \dfrac{7\pi}{4} < x \leq \dfrac{11\pi}{6}[/tex]

Assim, o conjunto solução da inequação dada é o seguinte:

[tex]S = \left\{x \in \mathbb{R} \, | \, \dfrac{\pi}{6} \leq x < \dfrac{\pi}{4} \,\,\, \lor \,\,\, \dfrac{3\pi}{4} < x \leq \dfrac{5\pi}{6} \,\,\, \lor \,\,\, \dfrac{7\pi}{6} \leq x < \dfrac{5\pi}{4} \,\,\, \lor \,\,\, \dfrac{7\pi}{4} < x \leq \dfrac{11\pi}{6}\right\}.[/tex]