1) т.к. 1.5π<α<2π
то cosα>0
⇒cosα = √(1-sin²α) = √(1-9/25) = 4/5
2)sin(π-α)=sinα
sin(π-α)=sinπ*cosα - sinα*cosπ = [т.к. sinπ=0 и cosπ=-1] = sinπ
3) sin(11π/4) = sin(3π/4) = √2/2
cos(13π/4) = cos(π/4) = √2/2
sin(-2.5π) = sin(-0.5π) = sin(-π/2) = -1
cos(-25π/3) = cos(25π/3) = cos(π/3) = 1/2
(√2/2 - √2/2) *(-1) * (1/2) = 0
4) cosα=-2/3
sinα = ±√(1-cos²α) = ±√(1-4/9) = ±√5/3
⇒|sinα|<1
√((1-sinα)/(1+sinα))=√((1-sinα)²/(1-sin²α))=√((1-sinα)²/cos²α)=|(1-sinα)|/|cosα|
√((1+sinα)/(1-sinα))=√((1+sinα)²/(1-sin²α))=√((1+sinα)²/cos²α)=
=|(1+sinα)|/|cosα|
|(1-sinα)|/|cosα| + |(1+sinα)|/|cosα| = (|1-sinα|+|1+sinα|)/|cosα| =
=(1-sinα+1+sinα)/|cosα| = 2/|cosα| = 2/ (2/3) = 3
Раскроем скобки:
(sina+sinB+sina-sinB)/(cosa+cosB+cosa-cosB)
Сокращаем и получается:
2sina/2cosa
sin/cos - это tg
=>2sina/2cosa=2tga
1)Разложить на множители(числитель)
2)Потом в числителе формула a²-b²
3)Потом n+3 сокращаем с нижним
n³-9n/n+3=n(n²-9)/n+3=n(n-3)(n+3)/n+3=n(n-3)=n²-3n
2x-5>0
x>2,5
x+3>0
x>-3
Ответ: Х принвдлежит (2,5;8)