Question

How do you solve \[\tan x = 1\] and find all exact general solutions?

Answer 1

Hint: In the given question, we have been asked to find the general value of a trigonometric function. Now, the argument of the given trigonometric function is a variable and is equal to a constant. First, we are going to write the angle which satisfies the given equation and lies in a table of values from \[0^\circ \] to \[90^\circ \], then we are going to find all the general solutions by using the formula of the periodicity of the given trigonometric function and then solving it.

Formula Used:
We are going to use the formula of periodicity of tangent function, which is:
\[\tan \left( {n\pi + \theta } \right) = \tan \left( \theta \right)\]

Complete step by step answer:
The given function is \[\tan x = 1\].
First, we have to find the value of the angle which satisfies the value.
And, we know,
$\Rightarrow$ \[\tan \left( {\dfrac{\pi }{4}} \right) = 1\]
Now, we just apply the formula of periodicity of tangent function, which is:
$\Rightarrow$ \[\tan \left( {n\pi + \theta } \right) = \tan \left( \theta \right)\]
Hence, the general solution of \[\tan x = 1\] is,
\[\tan \left( {n\pi + \dfrac{\pi }{4}} \right)\].

Additional Information:
In the given question, we applied the concept of periodicity of the tangent function, so it is necessary that we know the periodicity of each trigonometric function. The periodicity of sine, cosine, cosecant, and secant is \[2\pi \] while the periodicity of tangent and cotangent is \[\pi \].

Note: We just need to remember the periodicity of the trigonometric functions, like here, the tangent function has the periodicity of \[\pi \], i.e., every \[\tan \] value repeats after this interval, so, when we have to find the general solution of any such trigonometric function, with any given value, we just find the standard solution and apply the formula of periodicity for calculating the general solution.