Решение.
Если [tex]\bf cost=a[/tex] , то [tex]\bf t=\pm arccos\, a+2\pi k\ \ ,\ k\in \mathbb {Z}[/tex] .
[tex]\displaystyle \bf 1.\ \ cos3x=\frac{\sqrt3}{2}\\\\3x=\pm arccos\frac{\sqrt3}{2}+2\pi k\ \ ,\ k\in Z\\\\3x=\pm \frac{\pi}{6}+2\pi k\ \ ,\ k\in Z\\\\x=\pm \frac{\pi}{18}+\frac{2\pi k}{3}\ \ ,\ k\in Z[/tex]
[tex]\displaystyle \bf2.\ \ cos\frac{x}{4}=-\frac{\sqrt2}{2}\\\\\frac{x}{4}=\pm arccos(-\frac{\sqrt2}{2})+2\pi k\ \ ,\ k\in Z\\\\\frac{x}{4}=\pm (\pi -arccos\frac{\sqrt2}{2})+2\pi k\ \ ,\ k\in Z\\\\\frac{x}{4}=\pm (\pi -\frac{\pi}{4})+2\pi k\ \ ,\ k\in Z\\\\\frac{x}{4}=\pm \frac{3\pi }{4}+2\pi k\ \ ,\ k\in Z\\\\x=\pm 3\pi +8\pi k\ \ ,\ k\in Z[/tex]
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Решение.
Если [tex]\bf cost=a[/tex] , то [tex]\bf t=\pm arccos\, a+2\pi k\ \ ,\ k\in \mathbb {Z}[/tex] .
[tex]\displaystyle \bf 1.\ \ cos3x=\frac{\sqrt3}{2}\\\\3x=\pm arccos\frac{\sqrt3}{2}+2\pi k\ \ ,\ k\in Z\\\\3x=\pm \frac{\pi}{6}+2\pi k\ \ ,\ k\in Z\\\\x=\pm \frac{\pi}{18}+\frac{2\pi k}{3}\ \ ,\ k\in Z[/tex]
[tex]\displaystyle \bf2.\ \ cos\frac{x}{4}=-\frac{\sqrt2}{2}\\\\\frac{x}{4}=\pm arccos(-\frac{\sqrt2}{2})+2\pi k\ \ ,\ k\in Z\\\\\frac{x}{4}=\pm (\pi -arccos\frac{\sqrt2}{2})+2\pi k\ \ ,\ k\in Z\\\\\frac{x}{4}=\pm (\pi -\frac{\pi}{4})+2\pi k\ \ ,\ k\in Z\\\\\frac{x}{4}=\pm \frac{3\pi }{4}+2\pi k\ \ ,\ k\in Z\\\\x=\pm 3\pi +8\pi k\ \ ,\ k\in Z[/tex]