Question

решите срочно, на кону 4 ​

Answer 1

Ответ:

$3(sinx+cosx)=2sin2x\\\\Zamena:\ t=sinx+cosx\ \ ,\ \ t^2=sin^2x+cos^2x+2sinx\cdot cosx=1+sin2x\ ,\\\\sin2x=t^2-1\ \ ,\\\\3\, t=2\, (t^2-1)\ \ ,\ \ \ 2t^2-3\, t-2=0\ \ ,\ \ D=25\ ,\ t_1=-\dfrac{1}{2}\ ,\ t_2=2\ ,\\\\a)\ \ sinx+cosx=-\dfrac{1}{2}\ \Big|:\sqrt2\ \ ,\\\\\dfrac{1}{\sqrt2}\cdot sinx+\dfrac{1}{\sqrt2}\cdot cosx=-\dfrac{1}{2\sqrt2}\ \ \Rightarrow \ \ cos\dfrac{\pi }{4}\cdot sinx+sin\dfrac{\pi}{4}\cdot cosx=-\dfrac{1}{2\sqrt2}\ ,\\\\sin(x+\dfrac{\pi}{4})=-\dfrac{1}{2\sqrt2}\ ,$

$x+\dfrac{\pi}{4}=(-1)^{n}arcsin(-\dfrac{1}{2\sqrt2})+\pi n\ ,\ n\in Z\\\\x=-\dfrac{\pi}{4}\++(-1)^{n+1}arcsin\dfrac{1}{2\sqrt2}+\pi n\ ,\ n\in Z\\\\b)\ \ sinx+cosx=2\ \ \Rightarrow \ \ \ sin(x+\dfrac{\pi}{4})=2>1\ \ \Rightarrow \ \ x\in \varnothing \\\\Otvet:\ x=-\dfrac{\pi}{4}+(-1)^{n+1}arcsin\dfrac{1}{2\sqrt2}+\pi n\ ,\ n\in Z\ .$