Сначала:
arcCos√3/2 = π/6
arcSin√2/2 = π/4
arc tg √3 = π/3
Теперь решаем:
а) Cos(π + π/6) = -Сosπ/6 = -√3/2
б) Cos(π/2 - π/3) =Sinπ/3 = √3/2
в) 8Sin x = 7Cos x |: Сosx ≠0
8tg x = 7
tgx = 7/8
x = arctg(7/8) + πk, k∈Z
(x²-10x)²-3*(x²-10x)=108
(x²-10x)=t ⇒
t²-3t-108=0 D=441
t₁=12 x²-10x=12 x²-10x-12=0 x₁=5+√37 x₂=5-√37
t₂=-9 x²-10x=-9 x²-10x+9=0 D=64 x₃=9 x₄=1.
A²-6a+9-5a²-6a=-4a<span>²-11a+9
a=1/2 => -4*1\4-11*1\2+9=-1-5.5+9=2.5
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9у²-25 = 3²у²-5²= (3у)²-5²= (3у-5) (3у+5)
3x^2-10x+3=0
D=(-10)^2-4*3*3=64
x1=(10-8)/6=1/3
x2=(10+8)/6=3
3(x-1/3)(x-3)= (3x-1)(x-3)