<s>[INST] Trigonometry equation, odd-function. So I have the following equation:
$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$
It should be solved using the fact that Sin is an odd function, I can not really get the gripp of how and what I need to do? Any sugestions?
 [/INST] More unorthodoxly, we first take the midpoint of $x-\frac{\pi}{6}$ and $x + \frac{\pi}{4}$ before applying the angle sum and difference identities:
Let $A = x + \frac{\pi}{24}$ (the midpoint). Then we need
$$\sin\left(A - \frac{5\pi}{24}\right) + \cos \left(A + \frac{5\pi}{24}\right) = 0$$
$$\sin A\cos\frac{5\pi}{24} - \cos A\sin\frac{5\pi}{24} + \cos A\cos\frac{5\pi}{24} - \sin A\sin\frac{5\pi}{24} = 0$$
$$\sin A\left(\cos \frac{5\pi}{24} - \sin \frac{5\pi}{24}\right) + \cos A\left(-\sin\frac{5\pi}{24} + \cos \frac{5\pi}{24}\right) = 0$$
$$\left(-\sin\frac{5\pi}{24} + \cos \frac{5\pi}{24}\right)\left(\sin A + \cos A\right) = 0$$
so we need
$$\sin A + \cos A  = 0$$
$$\tan A = -1$$
$$\tan \left( x + \frac{\pi}{24}\right) = -1$$
which can be solved directly as per normal.
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