In each solution, n is an arbitrary integer.
1. Nothing fancy, just take the inverse tangent.
tan(x) = -5
x = arctan(-5) + nπ
x = -arctan(5) + nπ
2. Recall the identity cos(2x) = 2 cos²(x) - 1, then factorize:
cos(2x) - 3 cos(x) + 2 = 0
(2 cos²(x) - 1) - 3 cos(x) + 2 = 0
2 cos²(x) - 3 cos(x) + 1 = 0
(2 cos(x) - 1) (cos(x) - 1) = 0
2 cos(x) - 1 = 0 or cos(x) - 1 = 0
cos(x) = 1/2 or cos(x) = 1
[x = arccos(1/2) + 2nπ or x = -arccos(1/2) + 2nπ]
… or x = arccos(1) + 2nπ
x = π/3 + 2nπ or x = -π/3 + 2nπ or x = 2nπ
3. By definition, csc(x) = 1/sin(x) :
csc(2x) + √2 = 0
csc(2x) = -√2
1/sin(2x) = -√2
sin(2x) = -1/√2
2x = arcsin(-1/√2) + 2nπ or 2x = π - arcsin(-1/√2) + 2nπ
2x = -π/4 + 2nπ or 2x = 5π/4 + 2nπ
x = -π/8 + nπ or x = 5π/8 + nπ
4. More factorization:
2 sin²(x) - sin(x) = 0
sin(x) (2 sin(x) - 1) = 0
sin(x) = 0 or 2 sin(x) - 1 = 0
sin(x) = 0 or sin(x) = 1/2
x = arcsin(0) + 2nπ
… or [x = arcsin(1/2) + 2nπ or x = π - arcsin(1/2) + 2nπ]
x = 2nπ or x = π/6 + 2nπ or x = 5π/6 + 2nπ
5. Yet more factorization. Also recall that |cos(x)| ≤ 1 for all x.
4 cos²(x) - 8 cos(x) + 3 = 0
(2 cos(x) - 3) (2 cos(x) - 1) = 0
2 cos(x) - 3 = 0 or 2 cos(x) - 1 = 0
cos(x) = 3/2 or cos(x) = 1/2
The first case gives no solution since 3/2 > 1.
x = arccos(1/2) + 2nπ or x = -arccos(1/2) + 2nπ
x = π/3 + 2nπ or x = -π/3 + 2nπ
6. Recall the identity sin(a - b) = sin(a) cos(b) - cos(a) sin(b).
sin(2x) cos(x) - cos(2x) sin(x) = -√3/2
sin(2x - x) = -√3/2
sin(x) = -√3/2
x = arcsin(-√3/2) + 2nπ or x = π - arcsin(-√3/2) + 2nπ
x = -π/3 + 2nπ or x = 4π/3 + 2nπ