1/cos^2+3 tgx -5=0
(sin^ + cos^)/cos^2+3 tgx -5=0
tg^x +1 +3 tgx -5=0
tgx=t
t^2 +3t -4=0
D=25
t1=-4 ; tgx=-4 ; x1=2пи*n -19/45*пи , n E Z
t2=1 ; tgx=1 ; x2= пи*n + пи/4 , n E Z
ОТВЕТ
x1=2пи*n -19/45*пи , n E Z
x2= пи*n + пи/4 , n E Z
1) (-1)^5= -1
2) 3^4= 81
3) (-2,7)^4= 53,1441
4) (-2)^3= -8
5) (-5)^0= 1
6) (-7)^8= 5764801
4-1-5-3-2-6
<span>Sin3x+Sin2x+Sinx=0
2Sin2xCosx + Sin2x = 0
Sin2x(2Cosx +1) = 0
Sin2x = 0 или 2Cosx +1 = 0
2x = </span>πn , n ∈Z Cosx = -1/2
<span>x = </span>πn/2, n ∈Z x = +-arcCos(-1/2) + 2πk , k ∈Z
x = +-2π/3 + 2πk , k ∈Z <span>
</span>