Question

2sin(2x+pi/4)+корень из 2=0

Answer 1

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