2cosx - cos²x = 0
cosx(2 - cosx) = 0
cosx = 0
x = π/2 + πn, n ∈ Z
2 - cosx = 0
cosx = 2 - нет корней
Ответ: x = π/2 + πn, n ∈ Z.
B^2/(b^2+bc)=
b^2/b(b+c)=b/(b+c)
3sin²x+6sinxcosx-2sin²x-2cos²x-5cos²x+5sin²x=0/cos²x
6tg²x+6tgx-7=0
tgx=a
6a²+6a-7=0
D=36+168=204
a1=(-6-2√51)/12=-1/2-√51/6⇒tgx=-1/2-√51/6⇒x=-arctg(1/2+√51/6)+πn,n∈z
a2=-1/2+√51/6⇒tgx=-1/2+√61/6⇒x=arctg(√51/6-1/2)+πk,k∈z