<span>N 122.
</span>1)
sin(x+60°)cos(x+30°) =1/2 ;
(1/2)*(sin(x+60°+x+30°) + sin(x+<span>60°-x-30°) ) = 1/2 ;
</span>sin(90°+2x) + sin<span>30°= 1 ;
</span>cos2x + 1/2 =1 ;
cos2x = <span>1/2 ;
2x = </span>±π/3 +2π*n , n∈ Z.
x = ±<span>π/6 +π*n , n∈ Z.
</span>ответ : ±<span>π/6 +π*n , n∈ Z. </span>
2)
sin3xsin<span>x +cos4x </span>=0 ; * * *cos4x=cos(3x+x) = cos3xcosx -sin3xsinx * * *
sin3xsinx +cos3xcosx -sin3xsin<span>x </span> <span>=0 ;
</span>cos3xcosx <span>=0 ;
</span>[ cos3x =0 ; cosx =0.⇔ [ 3x =π/2 +π*n ; x =π/2 +π*n , n∈ Z. ⇔
[ x = π/6 +(π/3)*n ; x =π/2 +π*n , n∈ Z. * * * π/2 +π*n ⊂ π/6 +(π/3)*n * * *
ответ : π/6 +(π/3)*n , <span> n ∈ Z .
</span>3)
cos6xcos3x =cos7xcos4x ;
(1/2) * ( cos(6x-3x) +cos(6x+3x) ) = (1/2) * ( <span>cos(7x-4x) +cos(7x+4x) ) ;
</span>cos3x +cos9x = cos3x +cos1<span>1x ;
</span>cos1<span>1x -cos9x =0 ;
</span>-2sin (11x -9x) /2 * sin(11x +9x) /2 =0
sinx * sin10x =0 ⇔ [ sin<span>x =0 ; sin10x =0 . </span>⇒ [ x =πn ; 10x =πn , n n <span>∈ Z .
</span>ответ : π*n /10 , <span> n ∈ Z .
</span>4)
12cos² (x/2) = 9 - 4cos(3x/2) * cos(x/2) ; * * * cos² (α/2) = (1+cosα) /2 * * *
12* (1+cosx) /2 = 9 - 4*(1/2)* (<span> cos(3x/2 -x/2) + </span> cos(3x/2+<span>x/2) )</span> ;
6 +6cosx =2cosx +2cos2x ;
cos2x - 2cosx -3 =0 ;
2cos²x -1 -2cosx -3 =0 ;
cos²x -cosx -2 =0 ; * * * замена : cosx = t ; -1 ≤ t ≤1 * * *
t² -t -2 =0 ;
t₁ = 2 > 1 не удовлетворяет
t ₂ = -1⇒ cosx = -1 ;
x = π +2πn , n ∈ Z .
ответ : π +2π*n , n∈ Z .
Вроде нормально) но это может быть несовсем точным, так что можешь списать у однаклассников)
а8 =25
а3 = a8 -5d
а13 = a8 +5d
а3+а13 = (a8 -5d) +(a8 +5d) = 2*a8 = 2*25 = 50
Способ подстановки:
х-у=1
х²-2у=26
у=х-1
х²-2х+2-26=0
х²-2х-24=0
D/4=1+24=25=5²
х1=1+5=6; у1=6-1=5; точка (6; 5)
х2=1-5=-4; у2=-4-1=-5; точка (-4; -5)
Ответ: (6;5) и (-4; -5).
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0,5х-1=у²
у+3х-7=0
3х-6у²-6=0
3х=7-у
7-у=3х
7-у-6у²-6=0
6у²+у-1=0
D=1+4*6=25=5²
у1=(-1+5)/12=1/3; 0,5х=у²+1=1/9+1; х1=2 2/9 точка (2 2/9; 1/3)
у2=(-1-5)/12=-0,5; 0,5х=0,25+1=1,25; х2=2,5 точка (2,5; -0,5)
Ответ: (2 2/9; 1/3) и (2,5; -0,5).