Question

ДАЮ 100 БАЛЛОВ. помогите пожалуйста решить задние в файле​

Answer 1

$\displaystyle\bf\\Cos\Big(2x-\frac{\pi }{4}\Big)=\frac{\sqrt{2} }{2}\\\\\\2x-\frac{\pi }{4} =\pm \ arcCos\frac{\sqrt{2} }{2}+2\pi n,n\in Z\\\\\\2x-\frac{\pi }{4}= \pm \ \frac{\pi }{4}+2\pi n,n\in Z\\\\\\\left[\begin{array}{ccc}2x=-\dfrac{\pi }{4} +\dfrac{\pi }{4}+2\pi n,n\in Z \\2x=\dfrac{\pi }{4}+\dfrac{\pi }{4} +2\pi n, n\in Z \end{array}\right\\\\\\\left[\begin{array}{ccc}2x=2\pi n,n\in Z\\2x=\dfrac{\pi }{2}+2\pi n,n\in Z \end{array}\right$

$\displaystyle\bf\\\left[\begin{array}{ccc}x=\pi n,n\in Z\\x=\dfrac{\pi }{4}+\pi n,n\in Z \end{array}\right \\\\\\n=0 \ \ \ \Rightarrow \ \ \ x=\frac{\pi }{4} +\pi \cdot 0=\frac{\pi }{4} \\\\\\Otvet \ : \ x= \ \frac{\pi }{4}$