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 спасибо! :)

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 

1) sint+sint=√22sint=√2sint=√2/2x= П/4+2ПК  x=-3П/4+2ПК2)sinx/3=-1/2x/3=-П/6+2ПК       х/3=-5П/6+2ПКх=-П/2+6ПК           х+-5П/2+6ПК5cos²x+6sinx-6=05(1-sin²x)+6sinx-6=05-5sin²x+6sinx-6=0-5sin²x+6sinx-1=05sin²x-6sinx+1=0sinx=t5t²-6t+1=0D=9-5=4t1=(3+2)/5=1  t2=(3-2)/5=1/5sinx=1                     sinx=1/5x=п/2+2ПК               х=(-1)^k arcsin(1/5)+Пк