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

A) 2cosx - 1 = 0
2cosx = 1
cosx = 1/2
x = ±π/3 + 2πn, n ∈ Z

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

А) 2cosx-1=0
cosx=1/2
x=-π/6+2πk U x=π/6+2πk,k∈z

Б) sin^2x+3sinxcosx+2cos^2x=0/cos²x
tg²x+3tgx+2=0
tgx=a
a²+3a+2=0
a1+a2=-3 U a1*a2=2
a1=-2⇒tgx=-2⇒x=-arctg2+πk,k∈z
a2=1π/4+πk,k∈z