Solve this trignometric /algebraic equation

$\begin{aligned} \sin x & = \cos 2x \\ \sin x & = 1 - 2\sin^2 x \\ \implies 2\sin^2 x + \sin x - 1 & = 0 \\ (2\sin x - 1)(\sin x + 1) & = 0 \end{aligned}$

$\implies \begin{cases} 2\sin x - 1 = 0 & \implies \sin x = \frac 12 & \implies x = \frac \pi 6 & \implies \tan \frac \pi 6 = \frac 1{\sqrt 3} \approx \boxed{0.577} \\ \sin x + 1 = 0 & \implies \sin x = -1 & \implies x = -\frac \pi 2 & \implies \tan \left(- \frac \pi 2\right) = - \infty \small \color{#D61F06} \text{ Infinite} \end{cases}$

$\sin x = \cos(2x) \Rightarrow \cos(\dfrac {π}{2} - x) = \cos(2x)$

$\dfrac {π}{2} - x = 2x \Rightarrow x = \dfrac {π}{6}$

Also, $\dfrac {π}{2} - x = -2x \Rightarrow x = -\dfrac {π}{2}$ .Since, $\cos(β) = \cos(-β)$

$\tan(-\dfrac {π}{2}) = -\infty$ ,which is negative. So, not the answer.

$\tan(\dfrac {π}{6}) = \dfrac {1}{\sqrt{3}} = \boxed{0.57735026919}$