Question

Number of solutions of the equation $\tan x + \sec x = 2\cos x,x \in \left[ {0,\pi } \right]$ is
$\left( a \right)0$
$\left( b \right)1$
$\left( c \right)2$
$\left( d \right)3$

Answer 1

Hint: In this particular question use the concept that tan x = (sin x/cos x) and sec x = (1/cos x) so use them and try to make a quadratic equation in sin x, then try to solve the quadratic equation using factorization or by quadratic formula so, use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation:
$\tan x + \sec x = 2\cos x,x \in \left[ {0,\pi } \right]$
Now as we know that $\tan x = \dfrac{{\sin x}}{{\cos x}},\sec x = \dfrac{1}{{\cos x}}$ so use these properties in the given equation we have,
$ \Rightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}} = 2\cos x$
$ \Rightarrow 1 + \sin x = 2{\cos ^2}x$
Now as we know that ${\cos ^2}x = 1 - {\sin ^2}x$ so we have,
$ \Rightarrow 1 + \sin x = 2\left( {1 - {{\sin }^2}x} \right)$
$ \Rightarrow 2{\sin ^2}x + \sin x - 1 = 0$
Now as we see that this is a quadratic equation in sin x so apply quadratic formula we have,
$ \Rightarrow \sin x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where, a = 2, b = 1, and c = -1 so we have,
$ \Rightarrow \sin x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4\left( 2 \right)\left( { - 1} \right)} }}{{2\left( 2 \right)}}$
$ \Rightarrow \sin x = \dfrac{{ - 1 \pm \sqrt 9 }}{4} = \dfrac{{ - 1 \pm 3}}{4}$
\[ \Rightarrow \sin x = \dfrac{{ - 1 - 3}}{4},\dfrac{{ - 1 + 3}}{4}\]
\[ \Rightarrow \sin x = - 1,\dfrac{1}{2}\]
So, when sin x = -1
$ \Rightarrow x = {\sin ^{ - 1}}\left( { - 1} \right) = {\sin ^{ - 1}}\left( {\sin \left( {2n + 1} \right)\dfrac{\pi }{2}} \right)$, where n = 1, 2, 3......
$ \Rightarrow x = \left( {2n + 1} \right)\dfrac{\pi }{2}$
$ \Rightarrow x = \dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2},.... > \pi $
So, $x \notin \left[ {0,\pi } \right]$ so this is not the solution.
Now when, $\sin x = \dfrac{1}{2}$
$ \Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{6}} \right) = {\sin ^{ - 1}}\left( {\sin \dfrac{{5\pi }}{6}} \right)$ in the interval $\left[ {0,\pi } \right]$
\[ \Rightarrow x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},x \in \left[ {0,\pi } \right]\]
So the given equation has two solutions in the interval $\left[ {0,\pi } \right]$.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the quadratic formula which is given as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, so first convert the equation into quadratic form as above and then apply the quadratic formula we will get the solutions for the sin x, then find the solutions for x as above then check which of the solutions are in the interval$\left[ {0,\pi } \right]$, which is the required answer.