Question

РЕШИТЕ УРАВНЕНИЕ

4 + 2sin2x - 5sinx - 5cosx=0

Answer 1

$4+2sin2x-5sinx-5cosx=0\\\\4+2\, sin2x-5(sinx+cosx)=0\\\\t=sinx+cosx\; ,\\\\t^2=(sinx+cosx)^2=sin^2x+cos^2x+2\, sinx\cdot cosx=1+sin2x\; ,\\\\sin2x=t^2-1\\\\4+2(t^2-1)-5t=0\\\\2t^2-5t+2=0\; ,\; \; D=9\; ,\; t_1=\frac{1}{2}\; ,\; t_2=2\\\\a)\; \; sinx+cosx=\frac{1}{2}\; |:\sqrt2\\\\\frac{1}{\sqrt2}\, sinx+\frac{1}{\sqrt2}\, cosx=\frac{1}{2\sqrt2}\\\\sin\frac{\pi }{4}\cdot sinx+cos\frac{\pi }{4}\cdot cosx=\frac{\sqrt2}{4}\\\\cos(x-\frac{\pi }{4})=\frac{\sqrt2}{4}\\\\x-\frac{\pi}{4}=\pm arccos\frac{\sqrt2}{4}+2\pi n,\; n\in Z$

$x=\frac{\pi }{4}\pm arccos\frac{\sqrt2}{4}+2\pi n,\; n\in Z\\\\b)\; \; sinx+cosx=2\; |:\sqrt2\\\\cos(x-\frac{\pi }{4})=\sqrt2>1\; \; \Rightarrow \; \; x\in \varnothing \; \; (-1\leq cosx\leq 1)\\\\Otvet:\; \; x=\frac{\pi }{4}\pm arccos\frac{\sqrt2}{4} +2\pi n,\; n\in Z$