1) sin²x - sinx = 2
sinx(sinx - 1) = 2
sinx = 2 or sinx = -1
Since sinx cannot be greater than 1, we have sinx = -1
x = -π/2 + 2πn, where n is an integer.

2) 2cos²x = 1 + sinx
2(1 - sin²x) = 1 + sinx
2 - 2sin²x = 1 + sinx
2sin²x + sinx - 1 = 0
(2sinx - 1)(sinx + 1) = 0
sinx = 1/2 or sinx = -1
x = π/6 + 2πn, 5π/6 + 2πn, -π/2 + 2πn, where n is an integer.

3) 4cos2x - sin2x = 0
4(2cos²x - 1) - 2sinxcosx = 0
8cos²x - 8 - 2sinxcosx = 0
8cos²x - 8 - 4sinx(√(1-cos²x)) = 0
cos²x = 1 - sin²x
8(1-sin²x) - 8 - 4sinx(√(sin²x)) = 0
8sin²x - 4sinx - 8 = 0
2sinx(4sinx - 2) = 8
sinx(2sinx - 1) = 4
sinx = 1/2 or sinx = -2
x = π/6 + 2πn, 5π/6 + 2πn, (no solution for sinx = -2)

4) sin²x - 5sinxcosx + 4cos²x = 0
(sinxcosx - 4cosx)(sinxcosx - 1) = 0
cosx(sin²x - 4) = 0
cosx(1 - sin²x) = 0
cosx(1 + sinx)(1 - sinx) = 0
cosx = 0 or sinx = -1 or sinx = 1
x = π/2 + πn, -π/2 + 2πn, 2πn, where n is an integer.

5) cosx + cos3x + cos2x = 0
cosx + (4cos³x - 3cosx) + 2cos²x - 1 = 0
4cos³x + 2cos²x - 2cosx - 1 = 0
(2cosx + 1)(2cos²x - cosx - 1) = 0
cosx = -1/2 or cosx = 1 or cosx = -1
x = 2π/3 + 2πn, 0 + 2πn, π + 2πn, where n is an integer.