sin2t=(square root of 2)cost
[0,360)
Is this your equation,
$\displaystyle sin(2t) = \sqrt{2}~cos(t)$
or is it
$\displaystyle sin^2(t) = \sqrt{2}~cos(t)$
For the first:
$\displaystyle 2~sin(t)~cos(t) = \sqrt{2}~cos(t)$
$\displaystyle 2~sin(t) = \sqrt{2}$
$\displaystyle sin(t) = \frac{\sqrt{2}}{2}$
so t = 45, 135.
For the second:
$\displaystyle sin^2(t) = \sqrt{2}~cos(t)$
$\displaystyle 1 - cos^2(t) = \sqrt{2}~cos(t)$
$\displaystyle cos^2(t) + \sqrt{2}~cos(t) - 1 = 0$
Using the quadratic formula:
$\displaystyle cos(t) = \frac{-\sqrt{2} \pm \sqrt{2 + 4\sqrt{2}}}{2}$
The only valid solution uses the "+" sign, so
$\displaystyle cos(t) = \frac{-\sqrt{2} + \sqrt{2 + 4\sqrt{2}}}{2}$
which, numerically, gives t = 58.826, 301.174.
-Dan