How do you find the angle between the vector and the x-axis?

Hint: Since we have been told to find the general angle made by any arbitrary vector with x-axis, we will first try to consider an arbitrary vector. After this we will make the use of the given x-axis to find out the direction. And then follow the general process of finding the angle using direction cosines.

Complete step-by-step answer:
Since we have to find the angle between a vector and x-axis, let us consider an arbitrary vector as $ {\text{\vec A = }}{{\text{A}}_x}\hat i + {{\text{A}}_y}\hat j + {{\text{A}}_z}\hat k $ respectively.
The magnitude of vector $ {\text{\vec A}} $ will be $ \left| {{\text{\vec A}}} \right| = \sqrt {{{\text{A}}_x}{\text{ + }}{{\text{A}}_y}{\text{ + }}{{\text{A}}_z}} $
since the x-axis is given to us we will find the direction of the unit vector.
Unit direction in x-axis = $ \vec x = \hat i $
We know that to find the angle between vector and x-axis we have to find $ \cos \theta $
we know that $ \cos \theta = \dfrac{{{\text{\vec A}}{\text{.}}\hat i}}{{\left| {{\text{\vec A}}} \right|.\left| {\vec x} \right|}} $
 $
   \Rightarrow \cos \theta = \dfrac{{\left( {{{\text{A}}_x}\hat i + {{\text{A}}_y}\hat j + {{\text{A}}_z}\hat k} \right).\hat i}}{{\sqrt {{{\text{A}}_x}{\text{ + }}{{\text{A}}_y}{\text{ + }}{{\text{A}}_z}} .1}} \\
   \Rightarrow \cos \theta = \dfrac{{\left( {{{\text{A}}_x}} \right)}}{{\sqrt {{{\text{A}}_x}{\text{ + }}{{\text{A}}_y}{\text{ + }}{{\text{A}}_z}} }} \;
  $
hence $ \theta = {\cos ^{ - 1}}\left( {\dfrac{{\left( {{{\text{A}}_x}} \right)}}{{\sqrt {{{\text{A}}_x}{\text{ + }}{{\text{A}}_y}{\text{ + }}{{\text{A}}_z}} }}} \right) $ is the angle between vector and x-axis.

Note: The above angle obtained is the general angle obtained between any arbitrary vector and x-axis . If a specific vector is given and you have been asked to find the direction between the given vector and the x-axis then the above generalized angle can be used by changing its respective values.