Используя метод вс аргумента,решите неравенство 1)sin2x+√3*cos2x<0
2)√3*cos2x+sin2x≥√3​

1)arcsin 0 =0
2)arccos 1= 0 ;
3)arcsin√2/2 =π/4 ;
4)arccos 3  не  существует угол косинус которой =3 ;
5)arcsin (-1) = -π/2 ;
6)arccos(-√3/2) = π -π/6 = 5π/6 ;
7)arctg 0 = 0 ;
8)arctg 1 =π/4 ;
9)arctg(-√3) = - π/3 ;
10)arcctg(-√3/3) = π -π/3= 2π/3 ;
11)arcsin(-1/2)+arccos 1 = -π/6 +0 = -π/6 ;
12) (arcsin -1)/2+ arccos 1 = -π/4+0= -π/4;
13)cos ( arccos 1) =1;
14)sin(arcsin√2/2) =√2/2 ;
15)arcsin (sin π/4) =arcsin(√2/2) =π/4 ;
16)arccos ( cos(-π/4))=arccos ( cos(π/4))=arccos (√2/2))=π/4 ;
 17)cos (arcsin(-1/3))=cos(arccos(√8/3)= √8/3 =2√2/3 ;
18)tg(arccos(-1/4)) =tq(arctq(-√15) = - √15;   1+tq²α= 1/cos²α
19)sin(arcctg(-2)) =sin(arcsin(1/√5)=1/√5 ;
20) arcsin(cos π/9) =arcsin(sin(π/2 - π/9))=arcsin(sin7π/18) =7π/18 .

Объяснение:

(n+6)²=n²+12n+36

(13h+1)²=169h²+26h+1

(4-3y)²=16-24y+9y²

(2k-8)²=4k²-32k+64

(3c+7d)²=9c²+42cd+49d²

(9a+t)²=81a²+18at+t²

(k-8)²=k²-16k+64

(5-7m)²=25-70m+49m²

(13p-3)²=169p²-96p+9

(2f-10a)²=4f²-40af+100f²

(-3h+7)²=9h²-42h+49

(-10x-y)²=100x²+20xy+y²

(с-10)²=c²-28c+100

(11х+4)²=121x²+88x+16

(6+2y)²=36+24y+4y²

(4k-3)²=16k²-24k+9

(3c+2d)²=9c²+12cd+4d²

(8х-3у)²=64x²-48xy+9y²

(3а-5в)²=9a²-30aв+25в²

(7с-2m)²=49с²-28сm+4m²

(в+8)²=a²+16a+64

(12h+2)²=144h²+48h+4

(5-2y)²=25-20y+4y²

(3k-4)²=9k²-24k+16

(2c+5d)²=4c²+20cd+25d²

(8a+2t)²=64a²+32at²+4t²