Question

How do you solve for x in `sin(210+x)-cos(120+x)=0`?

Answer 1

We assume that `210` and `120` and angles in degrees.
This equation can be transformed into identity `0=0`, which means that any real value of `x` is a valid solution. Here is why.

Let's use the known equalities that immediately follow from the definition of `sin(phi)` and `cos(phi)` for any angle `phi` using a unit circle (all angles are measured in degrees):
`sin(180+phi)=-sin(phi)`
`cos(90+phi)=-sin(phi)`

Applying these equalities to our equation, we obtain
`sin(210+x)−cos(120+x)=0`
`sin(180+30+x)−cos(90+30+x)=0`
`-sin(30+x)+sin(30+x)=0`
`0=0`
The fact that we came to an unconditional identity for any variable `x` signifies that any real value of `x` is a solution.