Finding exact solutions to $\sin\left(\frac{\pi}{3} - x\right) - \cos\left(\frac{\pi}{6} + x\right) $

Question

Find the exact solutions to $$\sin\left(\frac{\pi}{3} - x\right) - \cos\left(\frac{\pi}{6} + x\right) $$

It says to solve for exact values algebraically but I'm not sure where to start. I've been trying different ways but nothing makes sense.

Answer 1

Your expression is identically $0$. The first term is $\cos (\frac {\pi} 2- (\frac {\pi} 3 -x))=\cos (\frac {\pi} 6+x)$

Answer 2

$$\sin \left(\frac{\pi }{3}-x\right)+\cos \left(\frac{\pi }{6}+x\right)=0$$

We have, $\cos \left(\frac{\pi }{6}+x\right)$=$=\cos \left(\frac{\pi }{6}\right)\cos \left(x\right)-\sin \left(\frac{\pi }{6}\right)\sin \left(x\right)$

(Simplifying): $\cos \left(\frac{\pi }{6}\right)\cos \left(x\right)-\sin \left(\frac{\pi }{6}\right)\sin \left(x\right)$

$=\frac{\sqrt{3}}{2}\cos \left(x\right)-\sin \left(\frac{\pi }{6}\right)\sin \left(x\right)$

$=\frac{\sqrt{3}}{2}\cos \left(x\right)-\frac{1}{2}\sin \left(x\right)$

Also, $\sin \left(\frac{\pi }{3}-x\right)$=$=-\cos \left(\frac{\pi }{3}\right)\sin \left(x\right)+\cos \left(x\right)\sin \left(\frac{\pi }{3}\right)$

$=-\frac{1}{2}\sin \left(x\right)+\sin \left(\frac{\pi }{3}\right)\cos \left(x\right)$

$=-\frac{1}{2}\sin \left(x\right)+\frac{\sqrt{3}}{2}\cos \left(x\right)$

Therefore, $$-\frac{1}{2}\sin \left(x\right)+\frac{\sqrt{3}}{2}\cos \left(x\right)+\frac{\sqrt{3}}{2}\cos \left(x\right)-\frac{1}{2}\sin \left(x\right)=0$$

Now, $-\frac{1}{2}\sin \left(x\right)+\frac{\sqrt{3}}{2}\cos \left(x\right)+\frac{\sqrt{3}}{2}\cos \left(x\right)-\frac{1}{2}\sin \left(x\right)$

$=-\frac{1}{2}\sin \left(x\right)-\frac{1}{2}\sin \left(x\right)+\frac{\sqrt{3}}{2}\cos \left(x\right)+\frac{\sqrt{3}}{2}\cos \left(x\right)$

As, $\frac{\sqrt{3}}{2}\cos \left(x\right)+\frac{\sqrt{3}}{2}\cos \left(x\right)$ $=\cos \left(x\right)\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)$

And $-\frac{1}{2}\sin \left(x\right)-\frac{1}{2}\sin \left(x\right)$ $=\sin \left(x\right)\left(-\frac{1}{2}-\frac{1}{2}\right)$ $=-\sin \left(x\right)$

So we have equation as $$-\sin \left(x\right)+\sqrt{3}\cos \left(x\right)=0$$

$\frac{-\sin \left(x\right)+\sqrt{3}\cos \left(x\right)}{\cos \left(x\right)}=\frac{0}{\cos \left(x\right)}$

$\sqrt{3}-\frac{\sin \left(x\right)}{\cos \left(x\right)}=0$

$\sqrt{3}-\tan \left(x\right)=0$

$\tan \left(x\right)=\sqrt{3}$

$$x=\frac{\pi }{3}+\pi n$$