B6) (sin²α -cos²α)sin²α -cos²α)/(1-cos²α)+2ctq²α -(1+ctq²α) =(sin²α -cos²α)/sin²α+ctq²α-1 =1-ctq²α +ctq²α -1 =0.
B7) (28sinα +21ccosα)/(3cosα -4sinαα) =(28+21ctqα)/(3ctqα - 4) =(28+21*3/4)/(3*3/4-4) = -25.
C1) (sinα+cosα)² =(1/2)² ⇒1 +2sinα*cosα =1/4 ⇒sinα*cosα = -3/8.
C2) √8*arctq(-arcsin(1/3)) = -√8*arctq(arctq1/√8) =-√8*1/√8 = -1.
C3) sqrt(√2 - 2cosα);
√2 -2cosα ≥0 ⇒cosα ≤ √2/2 ⇒ 2π*k+π/4 ≤α≤(2π -π/4) +2π*k ;
2π*k+π/4 <α< 7π/4 +2π*k , α∈ [2π*k+π/4 ; 7π/4 +2π*k].
Выносим <span /> и 4 : x^2(x+2)-4(x+2) =0
(x^2-4)(x+2)=0
(x+2)^2(x-2)=0
x € <u>2,-2</u>
5 y - y = 2x - x + 2
4y - x = 2
X=1024 : 2
X=512
Ответ : х= 512