Question

Solve the following equation $\cot \left( {\dfrac{x}{2}} \right) = 1$ in the interval $\left[ {0,2\pi } \right]$ ?

Answer 1

Hint: We have given a trigonometric equation and, we have to find the values of $x$ for which the given trigonometric equation holds in the given interval.
The general solution for the trigonometric equation $\cot \alpha = \cot \theta $ is given as $\alpha = n\pi + \theta $ where $n \in I$ and $\theta \in \left( {0,\pi } \right)$ .

Complete step by step solution:
Given a trigonometric equation is $\cot \left( {\dfrac{x}{2}} \right) = 1$.
The value of $\cot \theta = 1$ for $\theta = \dfrac{\pi }{4}$ so we can write the above equation as
$\cot \left( {\dfrac{x}{2}} \right) = \cot \dfrac{\pi }{4}$

Now using the solution of trigonometric equation given in hint, we get
$ \Rightarrow \dfrac{x}{2} = n\pi + \dfrac{\pi }{4}$

Now multiplying the above equation by $2$ , we get
$2 \times \dfrac{x}{2} = 2 \times \left( {n\pi + \dfrac{\pi }{4}} \right)$
$ \Rightarrow x = 2n\pi + \dfrac{\pi }{2}$

Now we substitute the different values of $n$ and find those values, which lies in the given interval of $\left( {0,2\pi } \right)$ .
For $n = 0$ the value of $x$ is given as
$ \Rightarrow x = 2\left( 0 \right)\pi + \dfrac{\pi }{2}$
$ \Rightarrow x = \dfrac{\pi }{2}$ This lies in the given interval.
For $n = 1$ the value of $x$ is given as
$ \Rightarrow x = 2\left( 1 \right)\pi + \dfrac{\pi }{2}$
$ \Rightarrow x = \dfrac{{5\pi }}{2}$ This does not lie in the given interval.
For $n = - 1$ the value of $x$ is given as
$ \Rightarrow x = 2\left( { - 1} \right)\pi + \dfrac{\pi }{2}$
$ \Rightarrow x = \dfrac{{ - 3\pi }}{2}$ This does not lie in the given interval.
For higher values of $n$ also, the value does not exist.

Hence, the only possible value of $x$ is $\dfrac{\pi }{2}$.

Note: Remember that the value of $\cot \theta $ is not defined for $\theta = 0$ and $\theta = \pi $ . So in the domain of $\cot \theta $ , $0$ and $\pi $ are excluded.
When substituting the values of $n$ , substitute both positive and negative values.