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} $ .