# I have the following equation: \sin(x-\frac{\pi}{6})+\cos(x+\frac{\pi}{4})=0

I have the following equation:
$\mathrm{sin}\left(x-\frac{\pi }{6}\right)+\mathrm{cos}\left(x+\frac{\pi }{4}\right)=0$
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nick1337

$\mathrm{cos}\left(x+\frac{\pi }{4}\right)=\mathrm{sin}\left(\frac{\pi }{4}-x\right)$
and the equation turns to
$\mathrm{sin}\left(x-\frac{\pi }{6}\right)=\mathrm{sin}\left(x-\frac{\pi }{4}\right)$
i.e.
$2x-\frac{5\pi }{12}=\left(2k+1\right)\pi$