6+x)...

August 2022 1 9 Report

Выручите пожалуйста.
〖cos〗^3 x+〖cos〗^2 x=0
〖sin〗^2 x 〖cos〗^2 x+sinx cosx=0
2cosx cos4x+cosx-2cos4x-1=0
tg 3x/2 cos⁡(π/6+x)=cos⁡(π/6+x)

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$1)\ \ cos^3x+cos^2x=0\ \ \ \to \ \ \ cos^2x\cdot (cosx+1)=0\ ,\\\\cosx=0\ \ \to \ \ \ x=\dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\\\cosx=-1\ \ \to \ \ \ x=\pi+2\pi k\ ,\ k\in Z\\\\Otvet:\ \ x=\dfrac{\pi}{2}+\pi n\ ,\ x=\pi +2\pi k\ ,\ n,k\in Z\ .$

$2)\ \ sin^2x\cdot cos^2x+sinx\cdot cosx=0\ \ \to \ \ sinx\cdot cosx\cdot (sinx\cdot cosx+1)=0\ ,\\\\a)\ \ sinx\cdot cosx=0\ \ \to \ \ \dfrac{1}{2}\cdot sin2x=0\ \ ,\ \ sin2x=0\ \ ,\ \ 2x=\pi n\ ,\\\\x=\dfrac{\pi n}{2}\ ,\ n\in Z\\\\b)\ \ sinx\cdot cosx=-1\ \ \to \ \ \dfrac{1}{2}\cdot sin2x=-1\ \ ,\ \ sin2x=-2\ \ \to \ \ x\in \varnothing \\\\Otvet:\ \ x=\dfrac{\pi n}{2}\ ,\ n\in Z\ .$

$3)\ \ 2cosx\cdot cos4x+cosx-2cos4x-1=0\\\\cosx\cdot (2cos4x+1)-(2cos4x+1)=0\\\\(2cos4x+1)(cosx-1)=0\\\\a)\ \ cos4x=-\dfrac{1}{2}\ \ \to \ \ 4x=\pm \dfrac{2\pi }{3}+2\pi n\ ,\ \ x=\pm \dfrac{\pi}{6}+\dfrac{\pi n}{2}\ ,\ n\in Z\\\\b)\ \ cosx=1\ \ \to \ \ x=2\pi k\ ,\ k\in Z\\\\Otvet:\ x=\pm \dfrac{\pi}{6}+\dfrac{\pi n}{2}\ ,\ x=2\pi k\ ,\ n,k\in Z\ .$

$4)\ \ tg\dfrac{3x}{2}\cdot cos\Big(\dfrac{\pi }{6}+x\Big)=cos\Big(\dfrac{\pi }{6}+x\Big)\\\\ tg\dfrac{3x}{2}\cdot cos\Big(\dfrac{\pi }{6}+x\Big)-cos\Big(\dfrac{\pi }{6}+x\Big)=0\\\\cos\Big(\dfrac{\pi }{6}+x\Big)\cdot \Big(tg\dfrac{3x}{2}-1\Big)=0\\\\a)\ \ cos\Big(\dfrac{\pi}{6}+x\Big)=0\ \ \to \ \ \dfrac{\pi}{6}+x=\dfrac{\pi}{2}+\pi n\ \ ,\ \ x=\dfrac{\pi }{3}+\pi n\ ,\ n\in Z\\\\b)\ \ tg\dfrac{3x}{2}=1\ \ \to \ \ \dfrac{3x}{2}=\dfrac{\pi}{4}+\pi k\ ,\ \ x=\dfrac{\pi}{6}+\dfrac{2\pi k}{3}\ ,\ k\in Z$

$Otvet:\ x=\dfrac{\pi }{3}+\pi n\ ,\ x=\dfrac{\pi}{6}+\dfrac{2\pi k}{3}\ ,\ n,k\in Z\ .$

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