To find the values of x that satisfy the given inequalities, we can first find the restricted values of x that fall within the specified intervals.

1) sin x > √3/2
Since sin x is positive in the first and second quadrants, we have:

x ∈ (π/3, 2π/3) U (4π/3, 5π/3)

2) sin x < √3/2
Similarly, since sin x is negative in the third and fourth quadrants, we have:

x ∈ (2π/3, 4π/3)

3) cos x > -√3/2 and cos x < 1/2
This inequality is satisfied in the first and fourth quadrants. So we have:

x ∈ (π/6, 5π/6) U (5π/6, 11π/6)

4) tan x < -√3/3
This inequality is satisfied in the second and fourth quadrants. So we have:

x ∈ (5π/6, 7π/6) U (11π/6, 13π/6)

To find the intersection of all the intervals, we need to consider the common values of x that satisfy all the given inequalities. So the final interval that satisfies all the inequalities is:

x ∈ (5π/6, 7π/6)

Therefore, the solutions to the inequalities sin x > √3/2, sin x < √3/2, cos x > -√3/2, cos x < 1/2, and tan x < -√3/3 are x ∈ (5π/6, 7π/6).