Вариант 4 см.вложение СРОЧНО!!!

Question 1

Вариант 4

см.вложение

СРОЧНО!!!

Question 2

1a $x = \frac{\pi}{2} + 2\pi k$

1б $x= 2\pi k \pm \frac{\pi}{3}$

1в $x = \pi k - \frac{\pi}{6}$

2а $2\cos^2 x - \cos x - 1 =0\\ (2\cos x + 1)(\cos x - 1) =0\\ x = \frac{\pi k}{3}$

2б $3 \sin^2 - 2\cos x +2 = 0\\ 3 \cos^2 + 2\cos x - 5 = 0\\ (\cos x - 1)(3\cos x + 5) = 0\\ x =2\pi k$

3а $\sqrt{3}\sin x + \cos x = 0\\ \frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x = 0\\ \cos \frac{\pi}{6} \sin x + \sin \frac{\pi}{6} \cos x = 0\\ \sin (\frac{\pi}{6} + x) = 0\\ \frac{\pi}{6} + x = \pi k\\ x = \pi k - \frac{\pi}{6}$

3б $\sin^2 x - 2\sqrt{3}\sin x\cos x + 3\cos^2 x = 0\\ \cos^2 x(\tan^2 x - 2\sqrt{3}\tan x + 3) = 0\\ (\tan x - \sqrt{3})^2 = 0\\ \tan x = \pm \sqrt{3}\\ x = \pi k \pm \frac{\pi}{3}$

4a $x = (2k+1)\pi \pm \arccos 0.7$

4б $x = \pi k + (-1)^k \arcsin(\frac{1}{4})$

4в $x = \pi k + \arctan 5$

5а $\sin x - \cos x = -1\\ \frac{\sqrt{2}}{2}\sin x - \frac{\sqrt{2}}{2}\cos x = -\frac{\sqrt{2}}{2}\\ \cos \frac{\pi}{4} \sin x - \sin \frac{\pi}{4}\cos x = -\frac{\sqrt{2}}{2}\\ \sin (x-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\\ x - \frac{\pi}{4} = \pi k - (-1)^k\frac{\pi}{4}\\ x = \pi k +(1 - (-1)^k)\frac{\pi}{4}$

5б $\cos 4x - \sin^2 x = 1\\ 2\cos^2 2x - 1 - \sin^2 x = 1\\ 2(2\cos^2 x - 1)^2 - 1 + \cos^2 x - 1 = 1\\ 8\cos^4 x - 7\cos^2 x - 1 = 0\\ (\cos^2 x - 1)(8\cos^2 x +1)=0\\ \begin{cases} \cos^2 x = 1\\ \cos^2 x = -0.125 < 0 \end{cases}\\ \cos x = \pm 1\\ x = \pi k$

6а $x \in (\frac{7\pi}{6}+2\pi k;\frac{11\pi}{6}+2\pi k)$

6б $x \in (2\pi k-\frac{2\pi}{3};2\pi k+\frac{2\pi}{3})$

6в $x \in (\pi k-\frac{\pi}{2};\pi k +\frac{\pi}{3}]$

7 3 часа