# A basic problem with Solve

I have this equation:

4b*Cos[2t]-4a*Sin[2t]==4Cos[2t]+8Sin[2t]

Which I would like to solve. Without using mathematica, you can pretty easily see that a = -2 and b = 1, but when I solve it with mathematica it gives me various long results including sometimes tan, sec cot etc.

I can break it up into to parts like this:

Solve[4b*Cos[2t]==4Cos[2t],b]
Solve[-4a*Sin[2t]==+8Sin[2t],a]

However, the point with mathematica isn't to make everything manually, and I would hope there is a method to use, without manually editing the equations.

So my question is: How do I solve this for a and b, with the results of a = -2 and b = 1?

• You can try to use SolveAlways function.
– mmal
Apr 26 '13 at 15:41
• So in the first example you have 1 Equation with 2 variables, in the second example you have 2 equations with 2 variables. Don't you mean Solve[{[4b*Cos[2t]==4Cos[2t],-4a*Sin[2t]==+8Sin[2t]},{a,b}]?
– Sos
Apr 26 '13 at 15:42
• I'm guessing that it has to be solved for any t, so the (proper) approach is to write SolveAlways[ 4 b*Cos[2 t] - 4 a*Sin[2 t] == 4 Cos[2 t] + 8 Sin[2 t], t] or rather SolveAlways[4 b*Cos[2 t] - 4 a*Sin[2 t] == 4 Cos[2 t] + 8 Sin[2 t], {Sin[2 t], Cos[2 t]}]
– mmal
Apr 26 '13 at 15:46
• @mmal Thank you it works. Should I delete this question? Apr 26 '13 at 15:47
• No, don't delete it (but you may try to improve it to make it clearer) . Let @mmal post his answer. We don't see enough SolveAlways[] uses around. Apr 26 '13 at 15:49

since we know that $\sin$ and $\cos$ are orthogonal functions.