Question

срочно подробное решение)

$8cos^{4}x=11cos2x-1$
$tg^{2}x-2tgx+6ctgx=3$

Answer 1

$8cos^4x = 11cos2x - 1\ 8cos^4x = 11(2cos^2x -1) - 1\ 8cos^4x = 22cos^2x - 11 -1\ 8cos^4x - 22cos^2x + 12 = 0\ cos^2x = t\ 8t^2 - 22t + 12 = 0\ D = 484 - 4*8*12 = 100\ t_{1,2} = frac{22 pm 10}{-16} = -2; -frac{1}{8}.\ cos^2x = -2\ -cosx neq sqrt{2} \ cos^2x = -frac{1}{8}\ -cos^2x = frac{1}{8}\ -cosx = frac{1}{2sqrt{2}}\ cosx = -frac{1}{2sqrt{2}}\ x =pm arccos(-frac{1}{2sqrt{2}}) +2pi*k, kin Z$

 

$tg^2x - 2tgx + 6ctgx = 3\ tg^2x - 2tgx = 3 - 6ctgx\ tg^2x - 2tgx = 3(1- 2ctgx)\ tg^2x - 2tgx = 3(1- 2tg^{-1}x)\ tg^2x = 3\ tgx = sqrt{3}\ x = frac{pi}{3}+pi*k, kin Z$