Решите уравнение:
А) 2cosx-1=0
Б) sin^2x+3sinxcosx+2cos^2x=0

a) 2cosx - 1 = 02cosx = 1cosx = 1/2x = ±π/3 + 2πn, n ∈ Zб) sin²x + 3sinxcosx + 2cos²x = 0     |:cos²xtg²x + 3tgx + 2 = 0tg²x + 2tgx + tgx + 2 = 0tgx(tgx + 2) + (tgx + 2) = 0(tgx + 1)(tgx + 2) = 0tgx = -1                          или               tgx = -2x = -π/4 + πn, n ∈ Z      или               x = arctg(-2) + πk, k ∈ Z

А) 2cosx-1=0cosx=1/2x=-π/6+2πk U x=π/6+2πk,k∈zБ) sin^2x+3sinxcosx+2cos^2x=0/cos²xtg²x+3tgx+2=0tgx=aa²+3a+2=0a1+a2=-3 U a1*a2=2a1=-2⇒tgx=-2⇒x=-arctg2+πk,k∈za2=1π/4+πk,k∈z