Haha I've got this equation while trying to do your other question

What did you get for $\displaystyle \sin 2\theta=\cos 3\theta$ in terms of sin and cos ?

Something like $\displaystyle 2\cos\theta\sin\theta=4\cos^3\theta-3\cos\theta$ (1) isn't it ? (next time, it's better you write the result you got, so that we can check it

)

Now, let $\displaystyle \theta=\frac{\pi}{10}$, which is correct, since it satisfies the initial equation.

$\displaystyle \cos\theta\neq 0$, so you can divide both sides of (1) by $\displaystyle \cos\theta$ to get $\displaystyle 2\sin \frac{\pi}{10}=4\cos^2\frac{\pi}{10}-3$

Then just use the well-known identity $\displaystyle \cos^2 a+\sin^2 a=1$ to transform $\displaystyle \cos$ into $\displaystyle \sin$

That's all folks !