Помогите решить, вроде как с заменой переменой. Не упростить, а решить.

1)

2)

3)

4)

5)

1) sinx - (1 - sin^2 (x)) - sin^2 (x) = 0

sinx - 1 + sin^2 (x) - sin^2 (x) = 0

sinx = 1

x = 2πk, k∈Z

2) sinx = t ∈[-1; 1]

6t^2 + t - 1 = 0, D=1+4*6 = 25

t1 = (-1-5) / 12 = - 6/12 = - 1/2, sinx = - 0.5, x = π/6 + 2πk и x = 5π/6 + 2πk, k∈Z

t2 = (-1+5) / 12 = 4/12 = 1/3, sinx = 1/3, x = arcsin (1/3) + 2πk и x = π - arcsin (1/3) + 2πk, k∈Z

3) 1 - sin^2 (x) - 4sinx + 3 = 0

-sin^2 (x) - 4sinx + 4 = 0

sin^2 (x) + 4sinx - 4 = 0

sinx = t ∈[-1; 1]

t^2 + 4t - 4 = 0, D=16+16=32

t1 = (-4-√32) / 2 < - 1

t2 = (-4+√32) / 2, x = arcsin ((-4+√32) / 2) + 2πk и x = π - arcsin ((-4+√32) / 2) + 2πk

4) sinx * (√3*sinx - 3cosx) = 0

sinx = 0, x = πk

tgx = √3, x = π/3 + πk

5) 2sin^2 (x) - √3*2sinx*cosx = 0

2sinx * (sinx - √3*cosx) = 0

sinx = 0, x = πk

tgx = √3, x = π/3 + πk