# Solve for exact values algebraically

by Grimestock   Last Updated September 11, 2019 10:20 AM

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.

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

Kavi Rama Murthy
September 11, 2019 10:00 AM

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

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

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

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

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

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

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

Kumar Nilesh
September 11, 2019 10:15 AM