Решите пожалуйста... (10 класс)
sin [tex] \frac{2}{3} [/tex]x=[tex] \frac{1}{2} [/tex],
cos (2-3x)=[tex] \frac{ \sqrt{2} }{2} [/tex]
tg [tex] \frac{ \pi }{ x^{2} } [/tex]=[tex]- \frac{ \sqrt{3} }{3} [/tex]
sin [tex] \frac{2}{3} [/tex]x=[tex] \frac{1}{2} [/tex],
cos (2-3x)=[tex] \frac{ \sqrt{2} }{2} [/tex]
tg [tex] \frac{ \pi }{ x^{2} } [/tex]=[tex]- \frac{ \sqrt{3} }{3} [/tex]
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П - "пи"
1) 2x/3=(-1)^n *arcsin(1/2)+Пn, nєZ
x=(-1)^n *3/2*П/6+3Пn/2, nєZ
x=(-1)^n *П/4+3Пn/2, nєZ
2) 2-3х=±arccos(sqrt(2)/2)+2Пn, nєZ
х=±П/12+2/3-2Пn/3, nєZ
3) П/x^2=arctg(-sqrt(3)/3)+Пn, nєZ
П/х^2=-П/6+Пn, nєZ
1/х^2=-1/6+n, nєZ
1/х^2=(6n-1)/6
х^2=6/(6n-1)
x=±sqrt(6/(6n-1))