помогите во вложениях !очень нужно..... Created by Leno4kAngel. algebra-ru

помогите во вложениях !очень нужно...

Leno4kAngel @Leno4kAngel

July 2021 1 23 Report

помогите во вложениях !очень нужно....

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Alphaeus

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1-й рисунок:
$6a)\\ -\pi - arcsin(-1/2) + 2 \pi k < 3x < arcsin(-1/2) + 2\pi k\\ -\pi + \frac{\pi}{6} + 2\pi k < 3x < -\frac{\pi}{6} + 2\pi k\\ -\frac{5}6}\pi + 2\pik < 3x < -\frac{\pi}{6} + 2\pi k\\ -\frac{5}{18}\pi + \frac{2}{3}\pi k < x < -\frac{\pi}{18} + \frac{2}{3}\pi k$
$6b)\\ cos7xcos3x - sin3xsin7x$
$7)\\ 2 sin^2 x-5sinxcosx+2 cos^2 x=0\\ 2\frac{sin^2 x}{cos^2 x}-5\frac{sixcosx}{cos^2 x}+2\frac{cos^2 x}{cos^2 x}=0\\ 2tg^2 x - 5tgx + 2=0\\ \\ tgx=t\\ 2t^2 - 5t + 2=0\\ D= 5^2-4*2*2 =25-16 = 9\\ t_{1}=\frac{5+\sqrt{9}}{2*2} = \frac{5+3}{4} = 2\\ t_{2}=\frac{5-\sqrt{9}}{2*2} = \frac{5-3}{4} = \frac{1}{2}\\ \\ tgx=2\\ x_{1}=arctg2+\pi k\\ \\ tgx=1/2\\ x_{2}=arctg(1/2)+\pi k$

$8)\\\\1 + sin2x - sinx = cosx\\ 1 + sin2x = sinx + cosx\\ sin^2 x + cos^2 x + 2sinxcosx = cosx + sinx\\ (sinx+cosx)^2 = sinx + cosx\\ (sinx+cosx)^2 - (sinx + cosx) = 0\\ (sinx+cosx)*((sinx + cosx)-1) = 0\\ \\$

$> 2-й рисунок: 1) В 2) Б 3) Б 4) В 5) <img src=$

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