If $3 \tan \left(x-15^{\circ}\right)=\tan \left(x+15^{\circ}\right), 0
Given: $3 \tan \left(x-15^{\circ}\right)=\tan \left(x+15^{\circ}\right)$
$\Rightarrow \frac{\tan \left(x+15^{\circ}\right)}{\tan \left(x-15^{\circ}\right)}=3$
Applying componendo and dividendo, we have
$\frac{\tan \left(x+15^{\circ}\right)+\tan \left(x-15^{\circ}\right)}{\tan \left(x+15^{\circ}\right)-\tan \left(x-15^{\circ}\right)}=\frac{3+1}{3-1}$
$\Rightarrow \frac{\frac{\sin \left(x+15^{\circ}\right)}{\cos \left(x+15^{\circ}\right)}+\frac{\sin \left(x-15^{\circ}\right)}{\cos \left(x-15^{\circ}\right)}}{\frac{\sin \left(x+15^{\circ}\right)}{\cos \left(x+15^{\circ}\right)}-\frac{\sin \left(x-15^{\circ}\right)}{\cos \left(x-15^{\circ}\right)}}=\frac{4}{2}$
$\Rightarrow \frac{\sin \left(x+15^{\circ}\right) \cos \left(x-15^{\circ}\right)+\cos \left(x+15^{\circ}\right) \sin \left(x-15^{\circ}\right)}{\sin \left(x+15^{\circ}\right) \cos \left(x-15^{\circ}\right)-\cos \left(x+15^{\circ}\right) \sin \left(x-15^{\circ}\right)}=2$
$\Rightarrow \frac{\sin \left(x+15^{\circ}+x-15^{\circ}\right)}{\sin \left(x+15^{\circ}-x+15^{\circ}\right)}=2$
$\Rightarrow \frac{\sin 2 x}{\sin 30^{\circ}}=2$
$\Rightarrow \sin 2 x=2 \times \frac{1}{2}=1$ $\left(\sin 30^{\circ}=\frac{1}{2}\right)$
$\Rightarrow \sin 2 x=\sin 90^{\circ}$
$\Rightarrow 2 x=90^{\circ}$ $\left(0
$\Rightarrow x=45^{\circ}$