Помогите пожалуйста . Created by sqwozi. algebra-ru

Помогите пожалуйста ...

July 2022 1 6 Report

Помогите пожалуйста

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DK954

а) $\frac{1}{2}arccos\frac{1}{2}+arctg1 = \frac{1}{2}*\frac{\pi }{3}+\frac{\pi }{4} = \frac{\pi }{6} +\frac{\pi }{4} = \frac{2\pi }{12}+\frac{3\pi }{12} = \frac{5\pi }{12}$

б) $2arcsin\frac{\sqrt{3} }{2} +3arcctg(-\sqrt{3}) = 2*\frac{\pi }{3}+3*\frac{5\pi }{6}=\frac{2\pi }{3}+\frac{5\pi }{2} = \frac{4\pi }{6} + \frac{15\pi }{6} = \frac{19\pi }{6}$

в) $sin(arccos\frac{\sqrt{3} }{2}) = sin\frac{\pi }{6} = \frac{1}{2} = 0,5$

а) cos x = -1/2

x = ± arccos(-1/2) + 2πn

x = ± π - arccos 1/2 + 2πn

x = ± π - π/3 + 2πn

x = ± 2π/3 + 2πn, n∈Z

б)

$tg(-3x)=\frac{1}{\sqrt{3} } \\-tg3x = \frac{\sqrt{3} }{3} \\tg3x = -\frac{\sqrt{3} }{3} \\3x = arctg(-\frac{\sqrt{3} }{3})+\pi n\\3x = -arctg\frac{\sqrt{3} }{3} + \pi n\\3x = -\frac{\pi }{6} + \pi n\\x = -\frac{\pi }{6}*\frac{1}{3} +\pi n*\frac{1}{3} \\x = -\frac{\pi }{18}+\frac{\pi n}{3} \\x = \frac{5\pi }{18} +\frac{\pi n}{3}$ n∈Z

в)

$sin(x-\frac{\pi }{3} )=-1\\x-\frac{\pi }{3} = \frac{3\pi }{2}+2\pi n\\x = \frac{3\pi }{2} + \frac{\pi }{3} + 2\pi n\\x = \frac{9\pi }{6} + \frac{2\pi }{6} + 2\pi n\\x = \frac{11\pi }{6} + 2\pi n$ n∈Z

3 votes Thanks 1

sqwozi Огромное спасибо!

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