Question

How do you find all solutions of the equation $\sec x+\tan x=1$ in the interval $\left[ 0,2\pi \right)$?

Answer 1

Hint: In this problem we need to calculate all the solutions of the given equation which is $\sec x+\tan x=1$. For solving this we are going to use the trigonometric identity which is ${{\sec }^{2}}x-{{\tan }^{2}}x=1$. We can simplify this identity by using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Here we will use the given equation $\sec x+\tan x=1$ and calculate the value of $\sec x-\tan x$. From the values of $\sec x+\tan x$ and $\sec x-\tan x$ we will calculate the value of $\sec x$ by adding both the values. After that we will use the basic trigonometric definition of $\sec x$ which is $\sec x=\dfrac{1}{\cos x}$ to calculate the value of $\cos x$. From this we can write the general solution of the equation and from the general solution we can calculate the solutions which are in the given interval $\left[ 0,2\pi \right)$.

Complete step by step answer:
Given equation, $\sec x+\tan x=1$.
We have the trigonometric identity ${{\sec }^{2}}x-{{\tan }^{2}}x=1$. Applying the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the trigonometric identity, then we will get
$\left( \sec x+\tan x \right)\left( \sec x-\tan x \right)=1$
Substituting the value $\sec x+\tan x=1$ in the above equation and simplifying it, then we will have
$\begin{align}
  & 1\left( \sec x-\tan x \right)=1 \\
 & \Rightarrow \sec x-\tan x=1 \\
\end{align}$
We have the values $\sec x+\tan x=1$, $\sec x-\tan x=1$. Adding both the values to get the value of $\sec x$, then we will get
$\sec x+\tan x+\left( \sec -\tan x \right)=1+1$
Simplifying the above equation by using mathematical operations, then we will get
$\begin{align}
  & \sec x+\tan x+\sec x-\tan x=2 \\
 & \Rightarrow 2\sec x=2 \\
 & \Rightarrow \sec x=1 \\
\end{align}$
From basic definitions of trigonometry, we have the formula $\sec x=\dfrac{1}{\cos x}$. Substituting this value in the above equation, then we will get
$\begin{align}
  & \dfrac{1}{\cos x}=1 \\
 & \Rightarrow \cos x=1 \\
\end{align}$
We have the trigonometric value $\cos 0{}^\circ =1$. Substituting this value in the above equation, then we will have
$\cos x=\cos 0{}^\circ $
We can write the general solution of the equation $\cos x=\cos \theta $ as $x=2k\pi \pm \theta $, $k\in Z$. So, the general solution of the equation $\cos x=\cos 0{}^\circ $ will be
$\begin{align}
  & x=2k\pi \pm 0 \\
 & \Rightarrow x=2k\pi \\
\end{align}$, $k\in Z$.
Now substitute $k=0,1,2,...$ in the general solution to get the solutions of the given equation, then we will get
$x=0,2\pi ,4\pi ,...$
The solutions of the given equation $\sec x+\tan x=1$ lies in the interval $\left[ 0,2\pi \right)$ is only $0$.

Note: We can also plot the graph of the given equation and observe the solution of the given equation in the given interval. The graph of the given equation $\sec x+\tan x=1$ will be

From the above graph also, we can observe that the solution of the given equation in the given interval is $0$ only. In the interval they have included the value of $0$ but not the value of $2\pi $, so we are not considering the value of $2\pi $ as a solution in the given interval.