Question

2sin(2x+pi/4)+корень из 2=0

Answer 1

$2~sinBig(2x+dfrac{pi}{4}Big)+sqrt{2}=0\ \ 2~sinBig(2x+dfrac{pi}{4}Big)=-sqrt{2}~~~~~Big|:2\ \sinBig(2x+dfrac{pi}{4}Big)=-dfrac{sqrt{2}}{2}\ \ \ 1)~~2x+dfrac{pi}{4}=-dfrac{pi}{4}+2pi n\ \~~~~~2x=-dfrac{pi}{4}-dfrac{pi}{4}+2pi n\ \ ~~~~~2x=-dfrac{pi}{2}+2pi n~~~~~Big|:2\ \ ~~~~~boxed{boldsymbol{x_1=-dfrac{pi}{4}+pi n;~~~nin Z}}\ \ \ 2)~~2x+dfrac{pi}{4}=-dfrac{3pi}{4}+2pi k\ \~~~~~2x=-dfrac{pi}{4}-dfrac{3pi}{4}+2pi k\ \ ~~~~~2x=-pi +2pi k~~~~~Big|:2\ \ ~~~~~boxed{boldsymbol{x_2=-dfrac{pi}{2}+pi k;~~~k in Z}}$