How do you solve for $x$ in $\sin \left( 210+x \right)-\cos \left( 120+x \right)=0$?

Hint: We explain the process of finding values for associated angles. We find the rotation and the position of the angle for $\sin \left( 210+x \right)$ and $\cos \left( 120+x \right)$. We explain the changes that are required for that angle. Depending on those things we find the solution.

Complete step-by-step answer:
We need to find the ratio value for $\sin \left( 210+x \right)$ and $\cos \left( 120+x \right)$.
For general form of $\sin \left( x \right)$ and $\cos \left( x \right)$, we need to convert the value of x into the closest multiple of $\dfrac{\pi }{2}$ and add or subtract a certain value $\alpha $ from that multiple of $\dfrac{\pi }{2}$ to make it equal to x.
Let’s assume $x=k\times \dfrac{\pi }{2}+\alpha $, $k\in \mathbb{Z}$. Here we took the addition of $\alpha $. We also need to remember that $\left| \alpha \right|\le \dfrac{\pi }{2}$.
Now we take the value of k. If it’s even then keep the ratio as it is and if it’s odd then the ratio changes to sin ratio from cos and vice-versa.
Then we find the position of the given angle as quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis.
If the angel falls in the first or fourth quadrant for $\cos \left( x \right)$ then the sign remains positive but if it falls in the second or third quadrant then the sign becomes negative.
Similarly, if the angel falls in the first or second quadrant for $\sin \left( x \right)$ then the sign remains positive but if it falls in the third or fourth quadrant then the sign becomes negative.
Depending on the sign and ratio change the final angle becomes $\alpha $ from x.
For $\sin \left( 210+x \right)$, we can express as $\sin \left( 210+x \right)=\sin \left( 2\times \dfrac{\pi }{2}+30+x \right)$.
The value of k is even which means the trigonometric ratio remains $\sin \left( x \right)$.
The position of the angle is in the third quadrant. The angle completes the half-circle 1 times and then goes $\left( 30+x \right)$. Therefore, the sign becomes negative.
The final form becomes $\sin \left( 210+x \right)=\sin \left( 2\times \dfrac{\pi }{2}+30+x \right)=-\sin \left( 30+x \right)$.
For $\cos \left( 120+x \right)$, we can express as $\cos \left( 120+x \right)=\cos \left( 1\times \dfrac{\pi }{2}+30+x \right)$.
The value of k is odd which means the trigonometric ratio changes from $\cos \left( x \right)$ to $\sin \left( x \right)$.
The position of the angle is in the second quadrant. The angle crosses the first quadrant and then goes $\left( 30+x \right)$. Therefore, the sign becomes negative.
The final form becomes $\cos \left( 120+x \right)=\cos \left( 1\times \dfrac{\pi }{2}+30+x \right)=-\sin \left( 30+x \right)$.
$\begin{align}
  & \sin \left( 210+x \right)-\cos \left( 120+x \right)=0 \\
 & \Rightarrow -\sin \left( 30+x \right)+\sin \left( 30+x \right)=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
Therefore, for any value of $x$ the left-side of the equation becomes 0.
The solution for $\sin \left( 210+x \right)-\cos \left( 120+x \right)=0$ is $\forall x\in \mathbb{R}$.

Note: We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of $\pi $ instead of $\dfrac{\pi }{2}$. It makes the multiplied number always even. In that case we don’t have to change the ratio. If $x=k\times \pi +\alpha =2k\times \dfrac{\pi }{2}+\alpha $. Value of $2k$ is always even.