Question

How do you solve $\sec x.\csc x = 2\csc x$?

Answer 1

Hint: In order to solve the given equation, subtract both the sides by $2\csc x$, get an equation. Take common value out of the equation, equate with zero to find the critical points. This method of solving the equations to find a general equation is known as trigonometric equations.

Formula used:
If $\cos x = \cos \alpha $ then\[x = \pm \alpha + 2\pi k, k \in Z\]
If $\sin x = \sin \alpha $ then\[x = \pi k + {\left( { - 1} \right)^k},k \in Z\]
If $\tan x = \tan \alpha $ then\[x = \pm \alpha + \pi k, k \in Z\]

Complete step by step solution:
We are given an equation $\sec x.\csc x = 2\csc x$.
Subtracting both the side by $2\csc x$, we get:
$
  \sec x.\csc x - 2\csc x = 2\csc x - 2\csc x \\
  \sec x\csc x - 2\csc x = 0 \\
 $
We can see that $\csc x$ is common, so taking out the common value and we get:
$\csc x\left( {\sec x - 2} \right) = 0$
And, we obtained two separate equations, that can either be equal to zero, that means either $\csc x = 0$ or $\sec x - 2 = 0$.
Considering both equal to zero one by one to get the critical points:
For $\csc x = 0$:
Since, we know that $\csc x = \dfrac{1}{{\sin x}}$, so substituting this value in the above equation and we get:
$
  \csc x = 0 \\
 \Rightarrow \dfrac{1}{{\sin x}} = 0 \\
 $
Solving for $\sin x$:
$\sin x = \dfrac{1}{0}$, which is of no real value hence, indefinite and do not exist. Therefore, we cannot consider $\csc x = 0$.

Moving on to $\sec x - 2 = 0$:
Since, we know that $\sec x = \dfrac{1}{{\cos x}}$, substituting this value in the equation and we get:
$
  \sec x - 2 = 0 \\
  \dfrac{1}{{\cos x}} - 2 = 0 \\
 $
Adding both sides by $2$,
$
  \dfrac{1}{{\cos x}} - 2 + 2 = 2 \\
  \dfrac{1}{{\cos x}} = 2 \\
 \Rightarrow \cos x = \dfrac{1}{2} \\
 $
Since, we know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$. So, comparing it with the above equation, we get:
$
  \cos x = \dfrac{1}{2} \\
  \cos \dfrac{\pi }{3} = \dfrac{1}{2} \\
  \cos x = \dfrac{\pi }{3} \\
  \Rightarrow x = \dfrac{\pi }{3} \\
 $
And, we obtained our principal value \[\alpha = \dfrac{\pi }{3}\].
But, though it’s not given to find the particular equation, so we will go for finding the general equation and we know that general equation for $\cos x$ if, $\cos x = \cos \alpha $ then\[x = \pm \alpha + 2\pi k, k \in Z\], where $\alpha $ is the principal angle which is just an acute angle.
According to this, the general value for our equation $\cos x = \dfrac{1}{2}$ is \[x = \pm \dfrac{\pi }{3} + 2\pi k, k \in Z\].

Therefore, the solution obtained by solving $\sec x.\csc x = 2\csc x$ is \[x = \pm \dfrac{\pi }{3} + 2\pi k, k \in Z\].

Note:
We can also leave the steps after finding the principal value but it’s good to find the general solution for convenience, as because the principal value is for a single quadrant whereas in general solution, we can get the solution for any quadrant by substituting the values we need.