tan(θ) = cot(30°+5θ)

Since tan(θ) = cot(90°-θ)

and you only want one value of θ,

Replace tan(θ) by cot(90°-θ)

cot(90°-θ) = cot(30°+5θ)

     90°-θ = 30°+5θ
       60° = 6θ
       10° = θ

[To get all the solutions, since tangent and cotangent
have period 180°, add n*180° to 90°-θ 

cot(90°-θ+n*180°) = cot(30°+5θ)

90°-θ+n*180° = 30°+5θ
  60°+n*180° = 6θ
   10°+n*30° = θ where n is any integer, positive, negative or zero. 
The solutions are given by this sequence:

..., -140°, -110°, -80°, -50°, -20°, 10°, 40°, 70°, 100°, 130°, 160°,... 

Edwin