Question

помогите решить, пожалуйста, алгебра 10 класс

Answer 1

$sinx+4sinx\, cosx+cosx=1\\\\(sinx+cosx)+4\cdot sinx\, cosx=1\\\\\star \; \; t=sinx+cosx\; ,\; \; t^2=\underbrace {sin^2x+cos^2x}_{1}+2\, sinx\, cosx\; \; \to \\\\t^2=1+2\, sinx\, cosx\; \; \to \; \; sinx\cdot cosx=\frac{t^2-1}{2}\; \; \star \\\\t+4\cdot \frac{t^2-1}{2}=1\; \; ,\; \; t+2t^2-2=1\; \; ,\; \; 2t^2+t-3=0\; ,\\\\D=1+24=25\; ,\; \; t_1=\frac{-1-5}{4}=-\frac{3}{2}\; \; ,\; \; t_2=\frac{-1+5}{4}=1\\\\a)\; \; sinx+cosx=-\frac{3}{2}\\\\sinx+sin(\frac{\pi}{2}-x)=-\frac{3}{2}\\\\2\, sin\frac{x+\frac{\pi}{2}-x}{2}\cdot cos\frac{x-\frac{\pi}{2}+x}{2}=-\frac{3}{2}$

$2\, sin\frac{\pi}{4}\cdot cos(x-\frac{\pi}{4})=-\frac{3}{2}\\\\\sqrt2\cdot cos(x-\frac{\pi}{4})=-\frac{3}{2}\\\\cos(x-\frac{\pi}{4})=-\frac{3\sqrt2}{4}\approx -1,06<-1\; \; \Rightarrow \; \; x\in \varnothing \\\\b)\; \; sinx+cosx=1\\\\\sqrt2\, cos(x-\frac{\pi}{4})=1\\\\cos(x-\frac{\pi}{4})=\frac{\sqrt2}{2}\\\\x-\frac{\pi}{4}=\pm \frac{\pi}{4}+2\pi n\; ,\; n\in Z\\\\x=\frac{\pi}{4}\pm \frac{\pi }{4}+2\pi n=\left [ {{\frac{\pi}{2}+2\pi n\; ,\; n\in Z} \atop {2\pi n\; ,\; n\in Z\quad }} \right. \\\\Otvet:\; \; x=\frac{\pi}{4}\pm \frac{\pi}{4}+2\pi n\; ,\; n\in Z\; .$