Question

How do you solve $\tan (x) = \dfrac{1}{2}$?

Answer 1

Hint: To solve the given question means to find the value of $x$. In order to find $x$ here, we will use inverse trigonometric functions. The inverse trigonometric functions are the inverse functions of the trigonometric functions which are used to obtain an angle from any of the angle’s trigonometric ratios. Here we will be using an inverse tan function to find the solution.

Complete step by step solution:
The trigonometric equation is $\tan (x) = \dfrac{1}{2}$.
On applying an inverse tan function, we get
$ \Rightarrow {\tan ^{ - 1}}(\tan x) = {\tan ^{ - 1}}(\dfrac{1}{2})$
$ \Rightarrow x = {\tan ^{ - 1}}(\dfrac{1}{2})$
$ \Rightarrow x = {26.565^o}$

Hence the solution of $\tan (x) = \dfrac{1}{2}$ is $x = {26.565^o}$.

Note: The domain of the inverse tan function is $( - \infty ,\infty )$which means domain contains all the real numbers and the range is $( - \dfrac{\pi }{2},\dfrac{\pi }{2})$ which means range contains all the angles between $ - \dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$ but not $ - \dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. In the question given above, when we used ${\tan ^{ - 1}}$, the domain for it was $\tan (x)$. The question itself states that $\tan (x) = \dfrac{1}{2}$, hence it passed the domain condition for the inverse tan function. Also \[{\tan ^{ - 1}}(\tan x) = x\] only when $x \in ( - \dfrac{\pi }{2},\dfrac{\pi }{2})$. Had $x$not assumed and lied between $( - \dfrac{\pi }{2},\dfrac{\pi }{2})$, we would not have been able to use \[{\tan ^{ - 1}}(\tan x) = x\].
One more thing that should also be kept in mind while using inverse trigonometric function (specifically inverse tan function for the context given) is that the expression${\tan ^{ - 1}}(x)$ is not equal to $\dfrac{1}{{\tan (x)}}$, which means $ - 1$ is not an exponent of $\tan (x)$ here.