I'm doing this practise exam question
' solve tan ((π/3) - x)= 0
the answer starts by swapping the inside function, but I don't understand how they have done this any help
tanπ -x=0
tanx-π =0
Hi, welcome to the forums! Please avoid double posting next time, if you need to modify your post you can press the "modify" button
Are you asking how they changed \(\tan \left(\frac{\pi }{3}-x\right)=0\) to \(\tan \left(x-\frac{\pi }{3}\right)=0\)?
If so, if you look at the unit circle we can see that \(\tan \left(x\right)\) is an odd function meaning that \(\tan \left(-x\right)=-\tan \left(x\right)\)
Therefore, \(\tan \left(\frac{\pi }{3}-x\right)=-\tan \left(-\left(\frac{\pi }{3}-x\right)\right)\) where you can simply divide -1 on each side to remove the negative equating to \(\tan \left(x-\frac{\pi }{3}\right)=0\).