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**TI-84** I'm a little confused after working your method out. Here's the way I did it.

$\displaystyle S = Sin\;t + cos\;2t$

$\displaystyle 0 = Sin\;t + cos\;2t$

$\displaystyle sin\;t + (1 - 2\;sin^2\;t) = 0$

$\displaystyle -2\;sin^2\;t + sin\;t + 1 = 0$

$\displaystyle (-sin\;t + 1) (2\;sin\;t + 1) = 0$

$\displaystyle -sin\;t = -1\Longrightarrow sin\;t = 1\Longrightarrow t = sin^{-1}(1)$

$\displaystyle 2\;sin\;t = -1\Longrightarrow sin\;t = \frac{-1}{2}\Longrightarrow t = sin^{-1}\left(\frac{-1}{2}\right)$

$\displaystyle t = 90\;or\; t = -30$

The $\displaystyle t = -30$ gets crossed out because its negative which leaves $\displaystyle t = 90$ as the final answer. Or did I mess up somewhere?