Question

Найти корени уровнения sin3x=корень2/2, на отрезке [0, 2pi];

Answer 1

$\sin3x=frac{sqrt2}2\3x=arcsinleft(frac{sqrt2}2right)=frac{pi}4+2pi n\x=frac{pi}{12}+frac{2pi}3n\xin[0,2pi]Rightarrow:\n=0quad x=frac{pi}{12}in[0,2pi]\n=1quad x=frac{pi}{12}+frac{2pi}3=frac{3pi}4in[0,2pi]\n=2quad x=frac{pi}{12}+frac{4pi}3=frac{17pi}{12}in[0,2pi]\n=3quad x=frac{pi}{12}+frac{6pi}3=frac{25pi}{12}notin[0,2pi]\x=frac{pi}{12}+frac{2pi}3n,quad n=0,1,2.$