sin(θ + 4°) = cos(θ + 6°)

Use the cofunction identity  cos(a) = sin{90°-a)

sin(θ + 4°) = sin[90° - (θ + 6°)]

sin(θ + 4°) = sin[90° - θ - 6°)]

sin(θ + 4°) = sin(84° - θ)

Then we use the fact that

if sin(α) = sin(β)

then one or both of these
hold

α = β + 360°n
α = (180°-β) + 360°n

for integer n

Using the first one
     α = β + 360°n
θ + 4° = 84° - θ + 360°n
    2θ = 80° + 360°n
     θ = 40° + 180°n 

letting n=0 give θ = 40°
letting n=1 give θ = 220°

Using the second one
     α = 180° - β + 360°n
θ + 4° = 180° - (84 - θ) + 360°n
θ + 4° = 180° - 84° + θ + 360°n
    4° = 96° + 360°n
  -92° = 360°n
   = n
    = n

That is extraneous because n is an integer.

So the only solutions between 0° and 360° are

θ = 40° and θ = 220°

and the general solution is  

θ = 40° + 180°n, for any integer n.

Edwin