Question

помогите во вложениях !очень нужно....

Answer 1

1-й рисунок: 
$6a)\\ -\pi - arcsin(-1/2) + 2 \pi k < 3x < arcsin(-1/2) + 2\pi k\\ -\pi + \frac{\pi}{6} + 2\pi k < 3x < -\frac{\pi}{6} + 2\pi k\\ -\frac{5}6}\pi + 2\pik < 3x < -\frac{\pi}{6} + 2\pi k\\ -\frac{5}{18}\pi + \frac{2}{3}\pi k < x < -\frac{\pi}{18} + \frac{2}{3}\pi k$
$6b)\\ cos7xcos3x - sin3xsin7x <= -1/2\\ cos(7x+3x)<= -1/2\\ cos 10x<= -1/2\\ 2\pi k + arccos(-1/2) < 10x < 2\pi (k+1) - arccos(-1/2)\\ 2\pi k + \frac{2}{3}\pi < 10x < 2\pi (k+1) - \frac{2}{3}\pi\\ \frac{1}{5}\pi k + \frac{1}{15}\pi < x < \frac{1}{5}\pi (k+1) - \frac{1}{15}\pi$
$7)\\ 2 sin^2 x-5sinxcosx+2 cos^2 x=0\\ 2\frac{sin^2 x}{cos^2 x}-5\frac{sixcosx}{cos^2 x}+2\frac{cos^2 x}{cos^2 x}=0\\ 2tg^2 x - 5tgx + 2=0\\ \\ tgx=t\\ 2t^2 - 5t + 2=0\\ D= 5^2-4*2*2 =25-16 = 9\\ t_{1}=\frac{5+\sqrt{9}}{2*2} = \frac{5+3}{4} = 2\\ t_{2}=\frac{5-\sqrt{9}}{2*2} = \frac{5-3}{4} = \frac{1}{2}\\ \\ tgx=2\\ x_{1}=arctg2+\pi k\\ \\ tgx=1/2\\ x_{2}=arctg(1/2)+\pi k$
 

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

$a) sinx + cosx = 0\\ tgx + 1 = 0\\ tgx = -1\\ x_{1} = arctg(1) + \pi n = \frac{\pi}{4} + \pi n\\ \\ b) sinx + cosx - 1 = 0\\ 2sin(x/2)cos(x/2) + cos^2(x/2) - sin^2(x/2) - sin^2(x/2) + cos^2(x/2) = 0\\ 2sin(x/2)cos(x/2) - 2sin^2(x/2)= 0\\ 2sin(x/2)*(cos(x/2) - sin(x/2))= 0\\ \\ b1) 2sin(x/2)= 0\\ x/2 = \pi n\\ x_{2}=2\pi n\\ \\ b2) cos(x/2) - sin(x/2)= 0\\ 1 - tg(x/2) = 0\\ tg(x/2)=1\\ x/2=\frac{\pi}{4} + \pi n\\ x_{3}=\frac{\pi}{2} + 2\pi n$

 

2-й рисунок:

1) В

2) Б

3) Б

4) В

5)

$<var>cosx - \sqrt{3}sinx=1\\ sinx = \frac{2tg\frac{x}{2}}{1+tg^2\frac{x}{2}}\\ cosx = \frac{1-tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}}\\ \frac{1-tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}} - \sqrt{3} * \frac{2tg\frac{x}{2}}{1+tg^2\frac{x}{2}} = 1\\ \frac{1-tg^2\frac{x}{2} - 2 \sqrt{3} tg\frac{x}{2}} {1+tg^2\frac{x}{2}}} = 1\\ 1-tg^2\frac{x}{2} - 2 \sqrt{3} tg\frac{x}{2} = 1+tg^2\frac{x}{2}}\\ 2tg^2\frac{x}{2} + 2 \sqrt{3} tg\frac{x}{2} =0\\ 2tg\frac{x}{2}*(tg\frac{x}{2} + \sqrt{3}) =0\\ a) \\ 2tg\frac{x}{2} =0\\ \frac{x}{2} = 0+ \pi n x_{1} = 2\pi n b)\\ tg\frac{x}{2} + \sqrt{3} =0\\ tg\frac{x}{2} =- \sqrt{3}\\ \frac{x}{2} =- \frac{\pi}{3} + \pi n\\ x =- \frac{2}{3}\pi + 2\pi n\\</var>$