Consider: \( f(x)=\sin \left(x+25^{\circ}\right) \cos 15^{\circ}-\cos \left(x+25^{\circ}\right) \sin 15^{\circ} \) Determine the general solution of \( f(x)=\tan 165^{\circ} \)

We start with \[ f(x)=\sin\left(x+25^\circ\right)\cos15^\circ-\cos\left(x+25^\circ\right)\sin15^\circ. \] Recognize the sine‐difference identity \[ \sin A\cos B-\cos A\sin B=\sin(A-B), \] with \[ A=x+25^\circ\quad\text{and}\quad B=15^\circ. \] Thus, \[ f(x)=\sin\bigl[(x+25^\circ)-15^\circ\bigr]=\sin(x+10^\circ). \] We now need to solve \[ \sin(x+10^\circ)=\tan165^\circ. \] Notice that \[ \tan165^\circ=\tan(180^\circ-15^\circ)=-\tan15^\circ. \] So the equation becomes \[ \sin(x+10^\circ)=-\tan15^\circ. \] For convenience, let \[ y=x+10^\circ. \] Then we have \[ \sin y=-\tan15^\circ. \] Because \(-\tan15^\circ\) is just a number in the interval \([-1,1]\) (since \(\tan15^\circ\approx0.268\)), the solution for \(y\) is given by the general solution to a sine‐equation: \[ y=\arcsin\bigl(-\tan15^\circ\bigr)+360^\circ\,k\quad \text{or}\quad y=180^\circ-\arcsin\bigl(-\tan15^\circ\bigr)+360^\circ\,k,\quad k\in\mathbb{Z}. \] Using the property \(\arcsin(-a)=-\arcsin(a)\) we can write \[ y=-\arcsin\bigl(\tan15^\circ\bigr)+360^\circ\,k\quad \text{or}\quad y=180^\circ+\arcsin\bigl(\tan15^\circ\bigr)+360^\circ\,k. \] Recalling that \(y=x+10^\circ\) we then have: \[ x+10^\circ=-\arcsin\bigl(\tan15^\circ\bigr)+360^\circ\,k \] or \[ x+10^\circ=180^\circ+\arcsin\bigl(\tan15^\circ\bigr)+360^\circ\,k. \] Solving for \(x\) gives: \[ x=-10^\circ-\arcsin\bigl(\tan15^\circ\bigr)+360^\circ\,k \] or \[ x=170^\circ+\arcsin\bigl(\tan15^\circ\bigr)+360^\circ\,k,\quad k\in\mathbb{Z}. \] Thus, the general solution is \[ \boxed{x=-10^\circ-\arcsin\bigl(\tan15^\circ\bigr)+360^\circ\,k\quad \text{or}\quad x=170^\circ+\arcsin\bigl(\tan15^\circ\bigr)+360^\circ\,k,\quad k\in\mathbb{Z}.} \]