If the matrices are multiplied

In[15]:= {{Cos[x], -Sin[x]}, {Sin[x], Cos[x]}} {{A, B}, {C, D}} {{Cos[x], Sin[x]}, {-Sin[x], Cos[x]}}
Out[15]= {{A Cos[x]^2, -B Sin[x]^2}, {-C Sin[x]^2, D Cos[x]^2}}

it seems evident that Reduce gives the correct, short answer (Mathematica 10.1)

In[13]:= Reduce[{{Cos[x], -Sin[x]}, {Sin[x], Cos[x]}} {{A, B}, {B, D}} {{Cos[x], Sin[x]}, {-Sin[x], Cos[x]}} == {{y, 0}, {0, z}}, {x, y, z}]
Out[13]= (B ==  0 || (C[1] \[Element] Integers && (x == 2 \[Pi] C[1] || x == \[Pi] + 2 \[Pi] C[1]))) &&
  y == A Cos[x]^2 && z == D Cos[x]^2

Solve does not miss parts of the solution

In[16]:= Solve[{{Cos[x], -Sin[x]}, {Sin[x], Cos[x]}} {{A, B}, {C,  D}} 
               {{Cos[x], Sin[x]}, {-Sin[x], Cos[x]}} == {{y, 0}, {0, z}}, {x, y, z}]

Out[16]= {{y -> A, z -> D, x -> ConditionalExpression[2 \[Pi] C[1], C[1] \[Element] Integers]}, 
          {y -> A, z -> D, x -> ConditionalExpression[\[Pi] + 2 \[Pi] C[1], C[1] \[Element] Integers]}}

as far as it presumes A, B, C, and D being unequal to zero:

In[17]:= Solve[{A Cos[x]^2 == y, -B Sin[x]^2 == 0, -C Sin[x]^2 == 0, D Cos[x]^2 == z}, {x, y, z}]
Out[17]= {{y -> A, z -> D, x -> ConditionalExpression[2 \[Pi] C[1],  C[1] \[Element] Integers]}, 
          {y -> A, z -> D, x -> ConditionalExpression[\[Pi] + 2 \[Pi] C[1], C[1] \[Element] Integers]}}