Решите уравнение 2(sin3x)^2+(cos3x)^2+sin3x=1

2Sin²3x + Cos²3x + Sin3x = 1

2Sin²3x + 1 - Sin²3x + Sin3x = 1

Sin²3x + Sin3x = 0

Sin3x(Sin3x + 1) = 0

$\left[\begin{array}{ccc}Sin3x=0\\Sin3x+1=0\end{array}\right\\\\\\\left[\begin{array}{ccc}Sin3x=0\\Sin3x=-1\end{array}\right\\\\\\\left[\begin{array}{ccc}3x=\pi n,n\in Z \\3x=-\frac{\pi }{2}+2\pi n,n\n Z\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\frac{\pi n}{3},n\in Z \\x=-\frac{\pi }{6}+\frac{2\pi n}{3},n\in Z \end{array}\right$