Question

помогите, пожалуйста!

Answer 1

$2Cos\Big(x+\dfrac{\pi }{4}\Big)=\sqrt{2}\\\\Cos\Big(x+\dfrac{\pi }{4}\Big)=\dfrac{\sqrt{2} }{2}\\\\x+\dfrac{\pi }{4}=\pm arc Cos\dfrac{\sqrt{2} }{2} +2\pi n,n\in Z\\\\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4} +2\pi n,n\in Z\\\\1)x+\dfrac{\pi }{4} =\dfrac{\pi }{4} +2\pi n,n\in Z\\\\x=\dfrac{\pi }{4} -\dfrac{\pi }{4} +2\pi n,n\in Z\\\\x=2\pi n,n\in Z\\\\0\leq2\pi n\leq2\pi \ |:2\pi\\\\0\leq n \leq 1\\\\n=0 \ \Rightarrow \ \boxed{x=0}\\\\n=1 \ \Rightarrow \ \boxed{x=2\pi}$

$2)x+\dfrac{\pi }{4}=-\dfrac{\pi }{4} +2\pi n,n\in Z\\\\x=-\dfrac{\pi }{4}-\dfrac{\pi }{4} +2\pi n,n\in Z\\\\x=-\dfrac{\pi }{2}+2\pi n,n\in Z\\\\0\leq -\dfrac{\pi }{2}+2\pi n\leq2\pi \ |\cdot \dfrac{2}{\pi }\\\\0\leq-1+4n\leq 4\\\\1\leq4n\leq5\\\\0,25\leq n \leq 1,25 \\\\n=1 \ \Rightarrow \ x=-\dfrac{\pi }{2}+2\pi=\boxed{\dfrac{3\pi }{2}} \\\\Otvet:\boxed{0 \ ; \ 2\pi \ ; \ \dfrac{3\pi }{2}}$

Answer 2

Ответ:

$x_1=0;\;\;\;\; x_2=2\pi ;\;\ x_3=\frac{3\pi }{2}$

Объяснение:

$2cos(x+\frac{\pi }{4} )=\sqrt{2} ;\;\;\ [0;2\pi ]\\\\cos(x+\frac{\pi }{4} )=\frac{\sqrt{2} }{2} ; \;\;\\\\x+\frac{\pi }{4} =\frac{+}{}\frac{\pi }{4} +2\pi k,\\ \\x=\frac{+}{} \frac{\pi }{4} -\frac{\pi }{4} +2\pi k \\\\\ x_1=\frac{\pi }{4} -\frac{\pi }{4} +2\pi k=0+2\pi k;\;\;\ k=0;1\;\;\ x_1=0;\;\ x_2=2\pi ;\\\\x_3=-\frac{\pi }{4} -\frac{\pi }{4} +2\pi k=-2*\frac{\pi }{4} +2\pi k=-\frac{\pi }{2}+2\pi k; k=1 \;\ =>\\\\x_3=-\frac{\pi }{2}+2\pi =\frac{3\pi }{2}$