Question

Помогите пожалуйста

Answer 1

$sin(\frac{\pi }{4}-8x)=-\frac{\sqrt3}{2}\\\\\frac{\pi }{4}-8x=(-1)^{n}\cdot (-\frac{\pi }{3})+\pi n=(-1)^{n+1}\cdot \frac{\pi}{3}+\pi n\; ,\; n\in Z\\\\8x=\frac{\pi }{4}-(-1)^{n+1}\cdot \frac{\pi }{3}-\pi n\\\\x=\frac{\pi }{32}+(-1)^{n+2}\cdot \frac{\pi}{24}-\frac{\pi n}{8}\; ,\; n\in Z$

$2)\; \; (a+3)\cdot sinx=a-1\; ,\; \; a>0\; ,\\\\sinx=\frac{a-1}{a+3}\; \; ,\; \; a\ne -3\; ,\\\\-1\leq \frac{a-1}{a+3}\leq 1\; \; \Rightarrow \; \; \left \{ {{\frac{a-1}{a+3}\leq 1} \atop {\frac{a-1}{a+3}\geq -1}} \right. \; \; \left \{ {{\frac{a-1-a-3}{a+3}\leq 0} \atop {\frac{a-1+a+3}{a+3}\geq 0}} \right. \; \left \{ {{\frac{-2}{a+3}\leq 0} \atop {\frac{2(a+1)}{a+3}\geq 0}} \right. \\\\\left \{ {{a+3\geq 0} \atop {a\in (-\infty ,-3)\cup (-1,+\infty )}} \right. \; \left \{ {{a\geq 0\; \; ,\; \; a>0} \atop {a\in (-\infty ,-3)\cup (-1,+\infty )}} \right. \; \Rightarrow \; \; a\in (0,+\infty )$