Question

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

Answer 1

2Sin²3x + Cos²3x + Sin3x = 1

2Sin²3x + 1 - Sin²3x + Sin3x = 1

Sin²3x + Sin3x = 0

Sin3x(Sin3x + 1) = 0

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