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**Prove It** Actually

$\displaystyle \sin{(2t + t)} = \sin{2t}\cos{t} + \cos{2t}\sin{t}$

$\displaystyle = 2\sin{t}\cos{t}\cos{t} + (1 - 2\sin^2{t})\sin{t}$

$\displaystyle = 2\sin{t}\cos^2{t} + \sin{t} - 2\sin^3{t}$

$\displaystyle = 2\sin{t}(1 - \sin^2{t}) + \sin{t} - 2\sin^3{t}$

$\displaystyle = 2\sin{t} - 2\sin^2{t} + \sin{t} - 2\sin^3{t}$

$\displaystyle = 3\sin{t} - 4\sin^3{t}$.

So $\displaystyle \displaystyle 3\sin{t} - 4\sin^3{t} = \frac{\sqrt{3}}{2}$

$\displaystyle \displaystyle 0 = 4\sin^3{t} - 3\sin{t} + \frac{\sqrt{3}}{2}$.

Now try to solve.