Question

Number of distinct solutions of
$\sec x + \tan x = \sqrt 3 ,x \in \left[ {0,3\pi } \right]$ is

Answer 1

Hint : In this question, to find the distinct solutions of the equation $\sec x + \tan x = \sqrt 3$ , we need to substitute secx as $\dfrac{1}{{\cos x}}$ and tanx as $\dfrac{{\sin x}}{{\cos x}}$ . Now, simplify the equation and then divide the whole equation with two. Then use the formulas
$\Rightarrow \cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow \sin \dfrac{\pi }{6} = \dfrac{1}{2}$
$\Rightarrow \cos \dfrac{\pi }{3} = \dfrac{1}{2}$
After substituting these values, use the formula $\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)$ . Now we have cos terms on both sides of the equation. So, we can now easily find the solution for x.

Complete step-by-step answer :
In this question, we are given a trigonometric equation and are supposed to find its number of distinct solutions.
The given equation is: $\sec x + \tan x = \sqrt 3$ - - - - - - - - - - - - - - (1)
Now, for solving this equation, we are going to use some trigonometric relations.
First of all, we can write secx as the inverse of cosx that is $\dfrac{1}{{\cos x}}$ and we can write tanx as $\dfrac{{\sin x}}{{\cos x}}$ .
Therefore, equation (1) becomes

$$\Rightarrow \sec x + \tan x = \sqrt 3 \\\ \Rightarrow \dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}} = \sqrt 3 \\\$$

Now, take cos x as a common denominator and take it to the LHS of the equation.
Therefore, we get

$$\Rightarrow \dfrac{{1 + \sin x}}{{\cos x}} = \sqrt 3 \\\ \Rightarrow 1 + \sin x = \sqrt 3 \cos x \\\$$

$\Rightarrow \sqrt 3 \cos x - \sin x = 1$ - - - - - - - - (2)
Now, here we will be dividing equation (2) with 2. Therefore, equation (2) becomes
$\Rightarrow \dfrac{{\sqrt 3 }}{2}\cos x - \dfrac{1}{2}\sin x = \dfrac{1}{2}$- - - - - - - - (3)
Now, we know that $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$ and $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$ and $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ . Therefore, substituting these values
in equation (3), we get
$\Rightarrow \cos \dfrac{\pi }{6}\cos x - \sin \dfrac{\pi }{6}\sin x = \cos \dfrac{\pi }{3}$
Now, we know the formula $\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)$ .Therefore, we get
$\Rightarrow \cos \left( {x + \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{3}$
Now, we know that when $\cos x = \cos y$ , we can say that $x = y$ . Therefore, we get

$$\Rightarrow x + \dfrac{\pi }{6} = 2n\pi \pm \dfrac{\pi }{3} \\\ \Rightarrow x = 2n\pi \pm \dfrac{\pi }{3} - \dfrac{\pi }{6} \\\$$

$\Rightarrow x = 2n\pi + \dfrac{\pi }{3} - \dfrac{\pi }{6} \\\ \Rightarrow x = 2n\pi + \dfrac{\pi }{6} \\\$
And
$\Rightarrow x = 2n\pi - \dfrac{\pi }{3} - \dfrac{\pi }{6} \\\ \Rightarrow x = 2n\pi - \dfrac{\pi }{2} \\\$
Now, we have to find the solutions in the range $\left[ {0,3\pi } \right]$ .
For $n = 0$ :
$\Rightarrow x = 2n\pi + \dfrac{\pi }{6} = \dfrac{\pi }{6}$
$\Rightarrow x = 2n\pi - \dfrac{\pi }{2} = - \dfrac{\pi }{2}$
But, $- \dfrac{\pi }{2} \notin \left[ {0,3\pi } \right]$ .
For $n = 1$ :
$\Rightarrow x = 2n\pi + \dfrac{\pi }{6} = 2\pi + \dfrac{\pi }{6} = \dfrac{{13\pi }}{6}$
$\Rightarrow x = 2n\pi - \dfrac{\pi }{2} = 2\pi - \dfrac{\pi }{2} = \dfrac{{3\pi }}{2}$

For $n = - 1$ :
$\Rightarrow x = 2n\pi + \dfrac{\pi }{6} = - 2\pi + \dfrac{\pi }{6} = \dfrac{{ - 11\pi }}{6}$
But, $- \dfrac{{11\pi }}{6} \notin \left[ {0,3\pi } \right]$ .
$\Rightarrow x = 2n\pi - \dfrac{\pi }{2} = - 2\pi - \dfrac{\pi }{2} = \dfrac{{ - 5\pi }}{2}$
But, $\dfrac{{ - 5\pi }}{2} \notin \left[ {0,3\pi } \right]$ .
For $n = 2$
$\Rightarrow x = 2n\pi + \dfrac{\pi }{6} = 4\pi + \dfrac{\pi }{6}$
But, $4\pi + \dfrac{\pi }{6} \notin \left[ {0,3\pi } \right]$ .
$\Rightarrow x = 2n\pi - \dfrac{\pi }{2} = 4\pi - \dfrac{\pi }{2}$
But, $4\pi - \dfrac{\pi }{2} \notin \left[ {0,3\pi } \right]$ .
Hence, the only possible values of x are $\dfrac{\pi }{6},\dfrac{{13\pi }}{6},\dfrac{{3\pi }}{2}$ .
So, the correct answer is “ $\dfrac{\pi }{6},\dfrac{{13\pi }}{6},\dfrac{{3\pi }}{2}$ .”.

Note : We can also solve this question using the identity $1 + {\tan ^2}x = {\sec ^2}x$ .
$\Rightarrow {\sec ^2}x - {\tan ^2}x = 1 \\\ \Rightarrow \left( {\sec x - \tan x} \right)\left( {\sec x + \tan x} \right) = 1 \;$
Now, $\sec x + \tan x = \sqrt 3$ . Therefore
$\Rightarrow \left( {\sec x - \tan x} \right)\sqrt 3 = 1$
$\Rightarrow \sec x - \tan x = \dfrac{1}{{\sqrt 3 }}$ - - - - - - (3)
Adding equation (1) and (2),we get
$\underline \sec x + \tan x = \sqrt 3 \\\ \sec x - \tan x = \dfrac{1}{{\sqrt 3 }} \\\ \\\ 2\sec x = \sqrt 3 + \dfrac{1}{{\sqrt 3 }} \\\$
$\Rightarrow \dfrac{2}{{\cos x}} = \dfrac{4}{{\sqrt 3 }} \\\ \Rightarrow \cos x = \dfrac{{\sqrt 3 }}{2} \;$
Now, the value of cosx is positive only in the 1st and 4th quadrant. So, the values of x will be
$0 + \dfrac{\pi }{6},2\pi - \dfrac{\pi }{6},2\pi + \dfrac{\pi }{6}$
Therefore, the solutions for $\sec x + \tan x = \sqrt 3$ will be $\dfrac{\pi }{6},\dfrac{{13\pi }}{6},\dfrac{{3\pi }}{2}$ .