Question

Решить уравнения: cos (π/2 - t) - sin (π+t) = √2 sin x/3 = -1/2 5 cos^2 x + 6 sin x - 6 = 0 заранее, огромное ! : )

Answer 1

Cos (π/2 - t) - Sin (π+t) = √2
sint + sint = √2
2sint = √2
sint = √2/2
t = (-1)^(n)*arcsin(√2/2) + πn, n∈Z
t = (-1)^(n)*(π/4) + πn, n∈Z

2)  Sin x/3 = -1/2
x = (-1)^(n)*arcsin(-1/2) + πk, n∈Z
x/3 = (-1)^(n+1)*arcsin(1/2) + πk, k∈Z
x/3 = (-1)^(n+1)*(π/6) + πk, k∈Z
x = (-1)^(n+1)*(3π/6) + 3πk, k∈Z
x = (-1)^(n+1)*(π/2) + 3πk, k∈Z

3)  5 Cos^2 x + 6 Sin x - 6 = 0
5*(1 - sin^2x) + 6sinx - 6 = 0
5 - 5*(sin^2x) + 6sinx - 6 = 0
5*(sin^2x) - 6sinx + 1 = 0
D = 36 - 4*5*1 = 16
a)  sinx = (6 - 4)/10
sinx = 1/5
x = (-1)^(n)*arcsin(1/5) + πn, n∈Z
б)  sinx = (6 + 4)/10
sinx = 1
x = π/2 + 2πk, k∈Z 

Answer 2

1) sint+sint=√2

2sint=√2

sint=√2/2

x= П/4+2ПК  x=-3П/4+2ПК

2)sinx/3=-1/2

x/3=-П/6+2ПК       х/3=-5П/6+2ПК

х=-П/2+6ПК           х+-5П/2+6ПК

5cos²x+6sinx-6=0

5(1-sin²x)+6sinx-6=0

5-5sin²x+6sinx-6=0

-5sin²x+6sinx-1=0

5sin²x-6sinx+1=0

sinx=t

5t²-6t+1=0

D=9-5=4

t1=(3+2)/5=1  t2=(3-2)/5=1/5

sinx=1                     sinx=1/5

x=п/2+2ПК               х=(-1)^k arcsin(1/5)+Пк