Question

How do you solve $1 - {\cot ^2}x = 0$ and find all solutions in the interval $0 \leqslant x < 360$?

Answer 1

Hint : Here, in the given question, we are given a trigonometric equation $1 - {\cot ^2}x = 0$ and we need to find all solutions of it in the interval $0 \leqslant x < 360$. The solutions of a trigonometric equation for which $0 \leqslant x < 360$ are called principal solutions. The expression involving integer $'n'$ which gives all solutions of a trigonometric equation is called the general equation. First, we will solve the given trigonometric equation $1 - {\cot ^2}x = 0$. We will try to convert $1 - {\cot ^2}x = 0$ in $\cot \theta = \cot \alpha$ form, because the general solution for the trigonometric equations of the form $\cot \theta = \cot \alpha$, is $\theta = n\pi + \alpha$ where $n \in Z$. After this, we will substitute the value of $\alpha$ in $\theta = n\pi + \alpha$ and find the solutions.

Complete step-by-step answer :
Given, $1 - {\cot ^2}x = 0$
On shifting ${\cot ^2}x$ to RHS, we get
$\Rightarrow 1 = {\cot ^2}x$
It can also be written as:
$\Rightarrow {\cot ^2}x = 1$
On taking square root both sides, we get
$\Rightarrow \cot x = \pm 1$
As we know, $\cot \dfrac{\pi }{4} = 1$. Therefore, we get
$\Rightarrow \cot x = \cot \left( { \pm \dfrac{\pi }{4}} \right)$
Now, we have the given trigonometric in the form of $\cot \theta = \cot \alpha$.
The general solution of $\cot \theta = \cot \alpha$ is $\theta = n\pi + \alpha$, where $n$ is an integer.
For $\cot x = \cot \left( { \pm \dfrac{\pi }{4}} \right)$ we have $\alpha = \pm \dfrac{\pi }{4}$
For $\alpha = \dfrac{\pi }{4}$, we have
When $n = 0$, we get
$\Rightarrow x = 0 \times \pi + \dfrac{\pi }{4} = \dfrac{\pi }{4}$
When $n = 1$, we get
$\Rightarrow x = 1 \times \pi + \dfrac{\pi }{4} = \dfrac{{4\pi + \pi }}{4} = \dfrac{{5\pi }}{4}$
When $n = 2$, we get
$\Rightarrow x = 2 \times \pi + \dfrac{\pi }{4} = \dfrac{{8\pi + \pi }}{4} = \dfrac{{9\pi }}{4}$
For $\alpha = - \dfrac{\pi }{4}$, we have
When $n = 0$, we get
$\Rightarrow x = 0 \times \pi - \dfrac{\pi }{4} = - \dfrac{\pi }{4}$
When $n = 1$, we get
$\Rightarrow x = 1 \times \pi - \dfrac{\pi }{4} = \dfrac{{4\pi - \pi }}{4} = \dfrac{{3\pi }}{4}$
When $n = 2$, we get
$\Rightarrow x = 2 \times \pi - \dfrac{\pi }{4} = \dfrac{{8\pi - \pi }}{4} = \dfrac{{7\pi }}{4}$
As mentioned in the question find solutions in the interval $0 \leqslant x < 360$, therefore we will not include $- \dfrac{\pi }{4}$ and $\dfrac{{9\pi }}{4}$ in our answer.
Hence, solutions of the trigonometric equation $1 - {\cot ^2}x = 0$ in the interval $0 \leqslant x < 360$ are $\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}$.
So, the correct answer is “$\dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}$”.

Note : Remember that the solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. Here, in the given question we have found the solutions of the trigonometric equation of $\cot$ function. Similarly, we can find the solutions of $\sin e$, $\cos ine$, $\tan$, etc. using their equation of general solution.
For the equation $\sin \theta = \sin \alpha$, write $\theta = n\pi + {\left( { - 1} \right)^n}\alpha$, $n \in Z$ as the general solution.
For the equation $\cos \theta = \cos \alpha$, write $\theta = 2n\pi \pm \alpha$, $n \in Z$ as the general solution.
For the equation $\tan \theta = \tan \alpha$, write $\theta = n\pi \pm \alpha$, $n \in Z$ as the general solution.
Remember that the general solutions of $\sin e$ and $\cos ecant$ are the same, general solutions of $co\sin e$ and $secant$ are the same, and general solutions of $\tan gent$ and $\cot angent$ are the same.