Solution of the equation 33sin3⁡x+cos3⁡x+33sin⁡xcos⁡x=1

It is in the form of a3+b3+c3−3abc=0(a+b+c)a2+b2+c2−ab−bc−ca=0,(3sin⁡x)3+(cos⁡x)3+(−1)3−3(3sin⁡x)(cos⁡x)(−1)=03sin⁡x+cos⁡x−1=03sin⁡x+cos⁡x=1, divided by 2sin⁡x+π6=sin⁡π6x+π6=nπ+(−1)nπ6,n∈z