Дима. 5-Д, 10-И, 0 лишний наверно, 14-М, 1-А.
4sin^2x - 17sinx+4 = 0 1) sinx=t 4t^2-17t+4=0. D=15^2 x1,2= 0,25;4 sinx=0,25 x=(-1)^n*arcsin0,25 + 2пиk k принадлежит z
1) sin(π/3 - α) + cos(π/6 - α) = sin(π/3)*cosα - cos(π/3)*sinα + cos(π/6)*cosα + sin(π/6)*sinα =(√3/2*)cosα - (1/2)*sinα + (√3/2)*cosα + (1/2)*sinα =
= (√3)*cosα
2) cosx - √3sinx = 1 делим на 2
(1/2)*cosx - (√3/2)*sinx = 1/2
cos(π/3)cosx - sin(π/3)*sinx = 1/2
cos(π/3 + x) = 1/2
x + π/3 = (+ -)*arccos(1/2) + 2πn, n∈Z
x + π/3 = (+ -)*(π/3) + 2πn, n∈Z
x = (+ -)*(π/3) - π/3 + 2πn, n∈Z
3cos²x+3sin²x+2sinxcosx-4sin²x=0/cos²x
tg²x-2tgx-3=0
tgx=a
a²-2a-3=0
a1+a2=2 U a1*a2=-3
a1=-1⇒tgx=-1⇒x=-π/4+πn,n∈z
a2=3⇒tgx=3⇒x=arctg3+πk,k∈z