Question

Решите уравнение и найдите наименьший положительный корень.
sin(2х + /4 ) = √2/ 2

Answer 1

$Sin\Big(2x+\dfrac{\pi }{4}\Big)=\dfrac{\sqrt{2} }{2}\\\\\\\left[\begin{array}{ccc}2x+\dfrac{\pi }{4} =arcSin\dfrac{\sqrt{2} }{2}+2\pi n,n\in Z \\2x+\dfrac{\pi }{4} =\pi-arcSin\dfrac{\sqrt{2} }{2}+2\pi n,n\in Z\end{array}\right\\\\\\\left[\begin{array}{ccc}2x+\dfrac{\pi }{4} =\dfrac{\pi }{4} +2\pi n,n\in Z \\2x+\dfrac{\pi }{4} =\pi-\dfrac{\pi }{4} +2\pi n,n\in Z\end{array}\right$

$\left[\begin{array}{ccc}2x+\dfrac{\pi }{4} =\dfrac{\pi }{4} +2\pi n,n\in Z \\2x+\dfrac{\pi }{4} =\dfrac{3\pi }{4} +2\pi n,n\in Z\end{array}\right\\\\\\\left[\begin{array}{ccc}2x=\dfrac{\pi }{4} -\dfrac{\pi }{4} +2\pi n,n\in Z \\2x=\dfrac{3\pi }{4}-\dfrac{\pi }{4}+2\pi n,n\in Z\end{array}\right\\\\\\\left[\begin{array}{ccc}2x=2\pi n,n\in Z \\2x=\dfrac{\pi }{2}+2\pi n,n\in Z \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\pi n,n\in Z \\x=\dfrac{\pi }{4}+\pi n,n\in Z \end{array}\right$

$n=0 \ \Rightarrow \ \boxed{x=\frac{\pi }{4}}$