Question

Resolve a inequação:
$sin (2x)= \frac{1}{2}$ ∧ x∈ [$-\pi , \pi$ ]

Answer 1

espero ter te ajudado muito! foi isso que consegui ok bjs

Answer 2

Resposta:

$S=\{-\frac{\pi }{12} ,\frac{\pi }{12} \}$

Explicação passo-a-passo:

$sen2x= \frac{1}{2} ~ e ~x \in~[-\pi ,~\pi ]\\\\2x=\frac{\pi }{6}+2k\pi \\\\x=\frac{\pi }{12}+k\pi \\\\p/x=-2\implies x=\frac{\pi }{12} -2\pi =-\frac{23\pi }{12}~~n\tildeao~~rvese \\\\p/x=-1\implies x=\frac{\pi }{12}-\pi =\frac{-11\pi }{12} \\\\p/x=0\implies x=\frac{\pi }{12}+0.\pi =\frac{\pi }{12}\\\\p/x=1\implies x=\frac{\pi }{12}+\pi =\frac{13\pi }{12}~~n\tilde ao~~serve$