I try to find $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$.
Left side:
$\sin(4 \alpha + \frac{\pi}{6}) =$ $= \sin4\alpha \times \cos \frac{\pi}{6} + \cos 4\alpha \times \sin\frac{\pi}{6} =$
$= \sin4\alpha \times \frac{\sqrt{3}}{2} + \cos 4\alpha \times \frac{1}{2} = $
$= 2\sin2\alpha \times \cos2\alpha \times \frac{\sqrt{3}}{2} + (\cos^2 2\alpha - \sin^2 2\alpha) \times \frac{1}{2} = $
$= 2\sin\alpha \times \cos \alpha \times \cos2\alpha \times \sqrt{3} + \frac{\cos^2 2\alpha - \sin^2 2\alpha}{2} = $
$= 2\sin\alpha \times \cos \alpha \times (\cos^2\alpha - \sin^2\alpha) \times \sqrt{3} + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha) - \sin^2 2\alpha}{2} = $
$= 2\sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha) - (2\sin\alpha \times \cos\alpha)(2\sin\alpha \times \cos\alpha)}{2} = $
$= 2\sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha)}{2} - 2\sin^2\alpha \times \cos^2\alpha = $
$= 2 \times \sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{\cos^4 \alpha - (\sin^2\alpha \times \cos^2\alpha) - (\sin^2\alpha \times \cos^2\alpha) + \sin^4 \alpha}{2} - 2\sin^2\alpha \times \cos^2\alpha = $
$= 2 \sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2\alpha - \sin^2\alpha)^2}{2} - 2\sin^2\alpha \times \cos^2\alpha = $
$= 2\sqrt{3}(\sin\alpha \times \cos^3 \alpha - \sin^3\alpha \times \cos\alpha) + \frac{(1 - 2\sin^2\alpha)^2}{2} - 2\sin^2\alpha \times \cos^2\alpha = $
$= 2\sqrt{3}(\sin\alpha \times \cos^3 \alpha - \sin^3\alpha \times \cos\alpha) + \frac{1}{2}-2\sin^2\alpha+2\sin^4\alpha - 2\sin^2\alpha \times \cos^2\alpha = $
$= 2\sin\alpha(\sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha) = $
Right side:
$\sin(2\alpha + \frac{\pi}{5}) = \sin2x \times \cos\frac{\pi}{5} + \cos2\alpha \times \sin\frac{\pi}{5} = $
$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + (\cos^2\alpha - \sin^2\alpha)\sin\frac{\pi}{5} = $
$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + (1 - 2\sin^2\alpha)\sin\frac{\pi}{5} = $
$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + \sin\frac{\pi}{5} - 2\sin^2\alpha \times \sin\frac{\pi}{5} = $
$= 2\sin\alpha(\cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5})$
I can't get any further on either. Bringing them together I get:
$2\sin\alpha(\sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha) = 2\sin\alpha(\cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5})$
$\iff \sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha = \cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5}$
Where do I go from here? Om am I already dead wrong?