25*sin(t) = 22*sin(2t) - 18*cos(2t)

sin(2t) = 2*sin(t)*cos(t)

cos(2t) = 1 - 2*sin^2(t)

So

25*sin(t) = 44*sin(t)*cos(t) - 18 + 36*sin^2(t)

Now, the problem is that we need to get rid of the cos(t) term somehow. There is only one reasonable way to do this: we know that sin^2(t) + cos^2(t) = 1, so...

cos(t) = (+/-)sqrt{1 - sin^2(t)}

Thus:

25*sin(t) = (+/-)44*sin(t)*sqrt{1 - sin^2(t)} - 18 + 36*sin^2(t)

Do yourself a favor at this point, let x = sin(t). Then:

25x = (+/-)44x*sqrt{1 - x^2} - 18 + 36x^2

Isolate the square root term:

(+/-)44x*sqrt{1 - x^2} = -36x^2 + 25x + 18

Square both sides:

1936x^2*(1 - x^2) = (-36x^2 + 25x + 18)^2

1936x^2 - 1936x^4 = 1296x^4 - 1800x^3 - 671x^2 + 900x + 324

3232x^4 -1800x^3 - 2607x^2 + 900x + 324 = 0

There's nothing easy about solving this, AND we need to check each x back in the original problem:

25x = (+/-)44x*sqrt{1 - x^2} - 18 + 36x^2

Once that is done, the final solution is

t = asn(x)

I have attached a graph of the polynomial function. I'll let you do the approximations and get the solutions for t.

-Dan