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Determine as raízes da equação tg²x = tg x, se [0,π].

Resposta :

[tex]85 \times 3 = 78[/tex]

espero de ajudando

[tex]\displaystyle \sf tg^2(x) =tg(x) \ \ ; \ \ x\in [0,\pi] \\\\ tg^2(x)-tg(x)= 0 \\\\ tg(x) (tg(x)-1) = 0 \\\\ tg(x) = 0 \to x = 0 \ \ ou \ \ x = \pi \\\\ tg(x) -1 = 0 \to tg(x) = 1 \to x= \frac{\pi}{4} \\\\ Portanto \ as\ solu\c c\~oes \ s\~ao : \\\\ \huge\boxed{\sf x = \left\{0,\pi,\frac{\pi }{4}\right\} }\checkmark[/tex]

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