Question

How does one find $\cot \theta - 1 = 0$ with the interval $ \left[ {0,2\pi } \right)$ with radians in terms of $\pi$?

Answer 1

Hint: This problem deals with trigonometry, the angles of the trigonometric function, the inverse of the trigonometric functions. Here, the value of the trigonometric ratio, we have to find the value of $\theta$ for that particular given equation, where the value of $\theta$ should be within the interval of $ \left[ {0,2\pi } \right)$. Here the found value of $\theta$ should be in terms of radians not degrees.

Complete step-by-step answer:
Given that $\cot \theta - 1 = 0$, expressing it mathematically below:
$ \Rightarrow \cot \theta - 1 = 0$
The above equation should be within the interval $ \left[ {0,2\pi } \right)$, expressing it mathematically below:
$ \Rightarrow \cot \theta - 1 = 0$; $\theta \in \left[ {0,2\pi } \right)$
Consider the given equation:
$ \Rightarrow \cot \theta - 1 = 0$
Rearranging the equation, as given below:
$ \Rightarrow \cot \theta = 1$
$ \Rightarrow \dfrac{1}{{\tan \theta }} = 1$
Reciprocating the above equation, as given below:
$ \Rightarrow \tan \theta = 1$
$ \Rightarrow \theta = {\tan ^{ - 1}}\left( 1 \right)$
As $\tan \dfrac{\pi }{4} = 1$, hence the value of $\theta $ is given below:
$ \Rightarrow \theta = \dfrac{\pi }{4}$
But the general solution of $\theta $ would be, as given below:
$ \Rightarrow \theta = \left( {2n + 1} \right)\dfrac{\pi }{4}$
Here $n = 1,2,3.....$
$n$ is integer here.
As given that the value of $\theta $ should be within the interval $\left[ {0,2\pi } \right)$, which means that 0 is included and $2\pi$ is excluded from the interval.
Hence $2\pi$ cannot be in one of the solutions of $\theta $. But the value of $\theta $ can be greater than zero and less than $2\pi $, which is mathematically expressed below;
$ \Rightarrow 0 \leqslant \theta < 2\pi $
Now we obtained the general solution for $\theta $ as,
$ \Rightarrow \theta = \left( {2n + 1} \right)\dfrac{\pi }{4}$
Substituting the values of $n$ until the value of cannot be equal or greater to the angle $2\pi $.
For $n = 0;$
$ \Rightarrow \theta = \dfrac{\pi }{4}$
For $n = 1;$
$ \Rightarrow \theta = \dfrac{{3\pi }}{4}$
For $n = 2;$
$ \Rightarrow \theta = \dfrac{{5\pi }}{4}$
For $n = 3;$
$ \Rightarrow \theta = \dfrac{{7\pi }}{4}$
So we obtained 4 general solutions for $\theta $ which are : $\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}.$
But if we substitute the value of $\dfrac{{3\pi }}{4}$ and $\dfrac{{7\pi }}{4}$ in $\theta $, the resulting value of $\cot \theta $ is -1, but not 1.
Hence the solutions for the equation $\cot \theta = 1$, are :
$ \Rightarrow \theta = \dfrac{\pi }{4},\dfrac{{5\pi }}{4}$

$\therefore $There are 2 solutions of $\theta $ which are $\dfrac{\pi }{4},\dfrac{{5\pi }}{4}$.

Note: Here while finding the value of $\theta$, we should understand that only those values of $\theta$ are considered, when obtained values of $\theta$ are substituted back in the equation, the values of $\theta$ should satisfy the given equation. Else those are not the correct values of $\theta$ for the equation. That is the reason why $\dfrac{{3\pi }}{4}$ and $\dfrac{{7\pi }}{4}$ even though obtained, they are not correct values of $\theta$, because when they were substituted back in the equation, they did not satisfy the equation.