Question

Вариант 4 

см.вложение

СРОЧНО!!!

 

Answer 1

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

 

1б $x= 2pi k pm frac{pi}{3}$

 

1в $x = pi k - frac{pi}{6}$

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

7 3 часа