Solve for x: \[{{\tan }^{-1}}x={{\cos }^{-1}}\dfrac{5}{13}\] Solve for \[\left[ 0,2\pi \right].....\] How?

Hint: These types of problems are pretty straight forward and are very easy to solve. It is a very good demonstration and example for trigonometric equations and general values. For these type of problems what we first do is, convert both the sides of the equation to a common form, which means, we can convert both the sides of the equation to either $\sin $ or $\cos $ or $\tan $ or $\cot $ or $\sec $ or $\text{cosec}$ . After we have converted it in the required form, we need to check for the quadrant in which the right hand side value lies and then accordingly we need to manipulate the given equation.

Complete step by step answer:
Now, we start off with the given problem as,
We convert the right hand side of the equation to its tangent form. If we construct an imaginary right angled triangle, and we consider the base as \[5\] and the hypotenuse as \[13\], then we get, \[\cos \theta =\dfrac{5}{13}\] . We will then get the value of \[\tan \theta \] as \[\tan \theta =\dfrac{12}{5}\] . We can therefore replace \[{{\cos }^{-1}}\dfrac{5}{13}\] by \[{{\tan }^{-1}}\dfrac{12}{5}\] .
Now, equating both the sides of the equation, we write,
\[{{\tan }^{-1}}x={{\tan }^{-1}}\dfrac{12}{5}\]
Now, comparing both the sides of the equation, we get the value of $x$ as,
\[\Rightarrow x=\dfrac{12}{5}\]

Note: For these types of above given problems we need to be very careful while applying the properties of inverse functions. A negative sign can cause a problem in the solution, which we can easily handle using the quadrant in which the value lies. We also need to keep in mind all the trigonometric general values, which become handy at critical times while solving complex problems of similar kind.