Question

The total number of solution of equation $\left| \left. \cot x \right| \right.=\cot x+\dfrac{1}{\sin x}$, x belongs to $\left[ 0,3\pi \right]$ is equal to
(a) 3
(b) 2
(c) 1
(d) 0

Answer 1

Hint: We will apply the concept of modulus here which is carried out as if we have $\left| \left. a \right| \right.$ then the value of this modulus is equal to $a$ and $-a$. This is because the modulus contains both the positive and negative sides of a. We will also use $\cot x=\dfrac{\cos x}{\sin x}$ to convert cotangent term into simpler form. Also, the use of the formula of $\cos x=\cos y$ which results in $x=2n\pi \pm y$ in the equations will help to solve the question further.

Complete step-by-step solution -
We will first consider the given trigonometric expression $\left| \left. \cot x \right| \right.=\cot x+\dfrac{1}{\sin x}$ ...(i). As $\cot x$ is in modulus so, we will use the formula of $\left| \left. a \right| \right.$ resulting into $a$ and $-a$ therefore, we get $\left| \left. \cot x \right| \right.$ as both $\cot x$ and $-\cot x$. This splits the trigonometric equation (i) into $\cot x=\cot x+\dfrac{1}{\sin x}$ and $-\cot x=\cot x+\dfrac{1}{\sin x}$. Now, we will consider these two equations one by one. We will consider the equation $\cot x=\cot x+\dfrac{1}{\sin x}$ first. After cancelling cotangent we get,
$\begin{align}
  & \cot x=\cot x+\dfrac{1}{\sin x} \\
 & \Rightarrow 1=1+\dfrac{1}{\sin x} \\
 & \Rightarrow 1=1+\dfrac{1}{\sin x} \\
 & \Rightarrow 1-1=\dfrac{1}{\sin x} \\
 & \Rightarrow 0=\dfrac{1}{\sin x} \\
\end{align}$
By using the property of reciprocal, we have
$\begin{align}
  & 0=\dfrac{1}{\sin x} \\
 & \Rightarrow \dfrac{0}{1}=\dfrac{1}{\sin x} \\
 & \Rightarrow \dfrac{\sin x}{1}=\dfrac{1}{0} \\
 & \Rightarrow \sin x=\infty \\
\end{align}$
As we know that the limit of the values for $\sin x$ lies between 1 and – 1. Thus this equation is not valid here.
Now we will consider the other equation which is $-\cot x=\cot x+\dfrac{1}{\sin x}$. In this equation we will convert it into simpler form by substituting $\cot x=\dfrac{\cos x}{\sin x}$ in $-\cot x=\cot x+\dfrac{1}{\sin x}$ equation. After doing this we get,
$\begin{align}
  & -\cot x=\cot x+\dfrac{1}{\sin x} \\
 & \Rightarrow -\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin x}+\dfrac{1}{\sin x} \\
 & \Rightarrow -\dfrac{\cos x}{\sin x}=\dfrac{\cos x+1}{\sin x} \\
 & \Rightarrow -\dfrac{\cos x}{\sin x}=\dfrac{\cos x+1}{\sin x} \\
 & \Rightarrow -\cos x=\cos x+1 \\
 & \Rightarrow -\cos x-\cos x=1 \\
 & \Rightarrow -2\cos x=1 \\
 & \Rightarrow \cos x=-\dfrac{1}{2} \\
\end{align}$
Now we will apply the value of the term $\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$ in the above equation. Thus, we now have $\cos x=-\cos \left( \dfrac{\pi }{3} \right)$. As we know that the value of the cosine is negative in the second and third quadrants so, we will consider second quadrant first. Therefore, by using the formula $\cos \left( x \right)=\cos \left( \pi -x \right)$ we will get
$\begin{align}
  & \cos x=\cos \left( \pi -\dfrac{\pi }{3} \right) \\
 & \Rightarrow \cos x=\cos \left( \dfrac{3\pi -\pi }{3} \right) \\
 & \Rightarrow \cos x=\cos \left( \dfrac{2\pi }{3} \right) \\
\end{align}$
Now we will apply the formula of $\cos x=\cos y$ which results into $x=2n\pi \pm y$ in the equation therefore we now have $x=2n\pi \pm \dfrac{2\pi }{3}$. Here n = ..., -1, 0, 1, ... As we put the value of n = 0 this results into $x=2n\pi \pm \dfrac{2\pi }{3}$. So, by substituting the values of n = 0 we get,
$\begin{align}
  & x=2\left( 0 \right)\pi \pm \dfrac{2\pi }{3} \\
 & \Rightarrow x=\pm \dfrac{2\pi }{3} \\
\end{align}$
Now we get two values of x which are $x=+\dfrac{2\pi }{3}$ and $x=-\dfrac{2\pi }{3}$. Now, as we can clearly see that $x=+\dfrac{2\pi }{3}$ belongs to the closed interval $\left[ 0,3\pi \right]$ while $x=-\dfrac{2\pi }{3}$ does not.
Similarly, the value of the cosine is negative third quadrant therefore by using the formula $\cos \left( x \right)=\cos \left( \pi +x \right)$ we will get
$\begin{align}
  & \cos x=\cos \left( \pi +\dfrac{\pi }{3} \right) \\
 & \Rightarrow \cos x=\cos \left( \dfrac{3\pi +\pi }{3} \right) \\
 & \Rightarrow \cos x=\cos \left( \dfrac{4\pi }{3} \right) \\
\end{align}$
As, $x=2n\pi \pm \dfrac{4\pi }{3}$. So, n = 0 this results into,
$\begin{align}
  & x=2\left( 0 \right)\pi \pm \dfrac{4\pi }{3} \\
 & \Rightarrow x=\pm \dfrac{4\pi }{3} \\
\end{align}$
Here, only $x=+\dfrac{4\pi }{3}$ belongs to the closed interval $\left[ 0,3\pi \right]$.
Since, these values were till 2 cycles of $\pi $. Now we will take the value from the third cycle of $\pi $ which is given by 2$\pi $ by the value of the cosine is negative third quadrant therefore by using the formula $\cos \left( x \right)=\cos \left( 3\pi -x \right)$ we will get
$\begin{align}
  & \cos x=\cos \left( 3\pi -\dfrac{\pi }{3} \right) \\
 & \Rightarrow \cos x=\cos \left( \dfrac{9\pi -\pi }{3} \right) \\
 & \Rightarrow \cos x=\cos \left( \dfrac{8\pi }{3} \right) \\
\end{align}$
By the formula of $\cos x=\cos y$ which results into $x=2n\pi \pm y$ in the equation therefore we now have $x=2n\pi \pm \dfrac{8\pi }{3}$ and n = 0 results into,
$\begin{align}
  & x=2\left( 0 \right)\pi \pm \dfrac{8\pi }{3} \\
 & \Rightarrow x=\pm \dfrac{8\pi }{3} \\
\end{align}$
Clearly, only $x=+\dfrac{8\pi }{3}$ belongs to the closed interval $\left[ 0,3\pi \right]$ while $x=-\dfrac{8\pi }{3}$ does not.
Hence the required values are $x=+\dfrac{4\pi }{3},x=+\dfrac{2\pi }{3}$ and $x=+\dfrac{8\pi }{3}$.

Note: As we know that the value of the cosine is negative in the second and third quadrants also we have the closed interval $\left[ 0,3\pi \right]$ for x therefore we will have the values for x which are in second and third quadrant only. Alternatively, we can solve the equation $-\cot x=\cot x+\dfrac{1}{\sin x}$ like below, and we can solve as usual.
  $\begin{align}
  & -\cot x=\cot x+\dfrac{1}{\sin x} \\
 & \Rightarrow -\cot x-\cot x=\dfrac{1}{\sin x} \\
 & \Rightarrow -2\cot x=\dfrac{1}{\sin x} \\
\end{align}$