Respuesta :

mreijepatmat
2cos(x)+sin(2x) =0

sin(2x) = 2sin(x).cos(x), then:

2cos(x)+sin(2x) = 2cos(x)+2sin(x).cos(x) =0; factorize:

2cos(x)[1+sin(x)]= 0, that means:

2cos(x)=0 and → x = π/2 + 2nπ, and

1+sin(x)= 0 and sin(x)=-1 → x=3π/2 + 2nπ

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