Question

ПОМОГМТЕ ПОЖАЛУЙСТА СРОЧНООООООООООООООООООООО УМОЛЯЮЮ ВВААААСССССС

Answer 1

Ответ:

$1)\ \ 1-2cosx\cdot sinx+\underbrace{sinx+cosx}_{u}=0\ \ ,\\\\\star \ \ u=sinx+cosx\ ,\\\\u^2=\underbrace {sin^2x+cos^2x}_{1}+2sinx\cdot cosx\ \ ,\ \ sinx\cdot cosx=\dfrac{u^2-1}{2}\ \ \star \\\\1-(u^2-1)+u=0\ \ ,\ \ \ u^2-u-2=0\ \ ,\ \ u_1=-1\ ,\ u_2=2\ \ ,\\\\a)\ \ sinx+cosx=-1\ \ ,\ \ \ \sqrt2\, \Big(\dfrac{1}{\sqrt2}\cdot sinx+\dfrac{1}{\sqrt2}\cdot cosx\Big)=-1\\\\cos\dfrac{\pi}{4}\cdot sinx+sin\dfrac{\pi}{4}\cdot cosx=-\dfrac{1}{\sqrt2}\ \ ,\ \ \ \ sin\Big(x+\dfrac{\pi}{4}\Big)=-\dfrac{1}{\sqrt2}$

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

$Otvet:\ \ x=\dfrac{\pi}{4}\cdot \Big((-1)^{n+1}-1\Big)+\pi n=\left[\begin{array}{l}-\dfrac{\pi}{2}+2\pi k,\ n=2k\in Z\\\pi (2k+1)\ ,\ n=2k+1\in Z\end{array}\right$

$2)\ \ sin2x+sin^2x=4cos^2x\\\\2sinx\cdot cosx+sin^2x-4cos^2x=0\ |:cos^2x\ne 0\\\\tg^2x+2tgx-4=0\ \ ,\ \ \ t=tgx\ ,\ \ \ t^2+2t-4=0\ \ ,\ \ D/4=5\ ,\\\\t_{1,2}=-1-\sqrt5\ \ ,\ \ t_2=-1+\sqrt5\\\\a)\ \ tgx=-1-\sqrt5\ \ ,\ \ x=arctg(-1-\sqrt5)+\pi n\ ,\ n\in Z\\\\b)\ \ tgx=-1+\sqrt5\ \ ,\ \ x=arctg(-1+\sqrt5)+\pi k\ ,\ k\in Z\\\\Otvet:\ \ x=arctg(-1-\sqrt5)+\pi n\ \ ,\ \ x=arctg(-1+\sqrt5)+\pi k\ ,\ n,k\in Z\ .$