Respuesta :

talahamze253

sin(3)+cos(3)sin(3)tan(3)3=0=−cos(3)=−1=−4+,∈ℤ=−12+3
sin⁡(3x)+cos⁡(3x) =0 sin⁡(3x) =−cos⁡(3x) tan⁡(3x) =−1 3x =−
π
4
+kπ,k∈
Z
x =−
π
12
+
k
π
3


Since −<<
−
π
<
x
<
π
, −2≤≤3
−
2
≤
k
≤
3
. Thus, the solution set is
{−34,−512,−12,4,712,1112}

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