Question

The number of solutions of the equation \[\tan x + \sec x = 2\cos x\] lying in the interval \[\left[ {0,2\pi } \right]\] is
(a) Zero
(b) One
(c) Two
(d) Three

Answer 1

Hint: Here, we need to find the number of solutions of the given equation. First, we will rewrite the given equation in terms of sine and cosine of \[x\] using trigonometric ratios. Then, we will simplify the equation using trigonometric identities to get a quadratic equation. Finally, we will solve the quadratic equation to get the possible values of \[x\], and hence, the number of solutions of the given equation.

Formula Used:
 We will use the following formulas:
1.The tangent of an angle \[\theta \] can be written as \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
2.The secant of an angle \[\theta \] can be written as \[\sec \theta = \dfrac{1}{{\cos \theta }}\].
3.We know that the sum of the square of the sine and cosine of an angle \[\theta \] is equal to 1, that is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].

Complete step-by-step answer:
First, we will rewrite the given equation in terms of sine and cosine of \[x\].
The tangent of an angle \[\theta \] can be written as \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].
Thus, we get
\[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
The secant of an angle \[\theta \] can be written as \[\sec \theta = \dfrac{1}{{\cos \theta }}\].
Thus, we get
\[\sec x = \dfrac{1}{{\cos x}}\]
Substituting \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] and \[\sec x = \dfrac{1}{{\cos x}}\] in the equation \[\tan x + \sec x = 2\cos x\], we get
\[ \Rightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}} = 2\cos x\]
Adding the terms in the expression, we get
\[ \Rightarrow \dfrac{{1 + \sin x}}{{\cos x}} = 2\cos x\]
Multiplying both sides by \[\cos x\], we get
\[\begin{array}{l} \Rightarrow \left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right)\cos x = 2\cos x \times \cos x\\ \Rightarrow 1 + \sin x = 2{\cos ^2}x\end{array}\]
We know that the sum of the square of the sine and cosine of an angle \[\theta \] is equal to 1, that is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
Thus, we get
\[{\sin ^2}x + {\cos ^2}x = 1\]
Subtracting \[{\sin ^2}x\] from both sides of the equation, we get
\[\begin{array}{l} \Rightarrow {\sin ^2}x + {\cos ^2}x - {\sin ^2}x = 1 - {\sin ^2}x\\ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x\end{array}\]
Substituting \[{\cos ^2}x = 1 - {\sin ^2}x\] in the equation \[1 + \sin x = 2{\cos ^2}x\], we get
\[ \Rightarrow 1 + \sin x = 2\left( {1 - {{\sin }^2}x} \right)\]
Multiplying the terms using the distributive law of multiplication, we get
\[ \Rightarrow 1 + \sin x = 2 - 2{\sin ^2}x\]
Rewriting the expression, we get
\[ \Rightarrow 2{\sin ^2}x + \sin x + 1 - 2 = 0\]
Subtracting the terms in the expression, we get
\[ \Rightarrow 2{\sin ^2}x + \sin x - 1 = 0\]
Let \[\sin x = y\].
Thus, the equation becomes
\[ \Rightarrow 2{y^2} + y - 1 = 0\]
We can see that this is a quadratic equation.
We will solve this quadratic equation using splitting the middle term and factorisation to find the values of \[y\].
Splitting the middle term, we get
\[ \Rightarrow 2{y^2} + 2y - y - 1 = 0\]
Factorising the terms, we get
\[\begin{array}{l} \Rightarrow 2y\left( {y + 1} \right) - 1\left( {y + 1} \right) = 0\\ \Rightarrow \left( {2y - 1} \right)\left( {y + 1} \right) = 0\end{array}\]
Therefore, we get
\[ \Rightarrow 2y - 1 = 0\] or \[y + 1 = 0\]
Simplifying the expressions, we get
\[ \Rightarrow y = \dfrac{1}{2}\] or \[y = - 1\]
Substituting \[y = \sin x\] in the expressions, we get
\[ \Rightarrow \sin x = \dfrac{1}{2}\] or \[\sin x = - 1\]
We know that \[\sin x = \dfrac{1}{2}\] is possible when \[x = \dfrac{\pi }{6}\] or \[x = \pi - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6}\].
Also, \[\sin x = - 1\] is possible when \[x = \dfrac{{3\pi }}{2}\].
Thus, we get the values of \[x\] as
\[x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{3\pi }}{2}\]
However, if we substitute \[x = \dfrac{{3\pi }}{2}\] in the given equation \[\tan x + \sec x = 2\cos x\], we get
\[ \Rightarrow \tan \left( {\dfrac{{3\pi }}{2}} \right) + \sec \left( {\dfrac{{3\pi }}{2}} \right) = 2\cos \left( {\dfrac{{3\pi }}{2}} \right)\]
Here, \[\dfrac{{3\pi }}{2}\] is an odd multiple of \[\dfrac{\pi }{2}\].
Thus, the equation \[\tan \left( {\dfrac{{3\pi }}{2}} \right) + \sec \left( {\dfrac{{3\pi }}{2}} \right) = 2\cos \left( {\dfrac{{3\pi }}{2}} \right)\] is not possible because the tangent of any odd multiple of \[\dfrac{\pi }{2}\] is not defined.
Therefore, the value of \[x\] cannot be \[\dfrac{{3\pi }}{2}\].
Thus, we get
\[ \Rightarrow x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}\]
\[\therefore \] We get the solutions of the equation \[\tan x + \sec x = 2\cos x\] as \[\dfrac{\pi }{6},\dfrac{{5\pi }}{6}\].
Since there are only two solutions, the correct option is option (c).

Note: We have used the distributive law of multiplication to multiply 2 by \[1 - {\sin ^2}x\]. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
We can make a mistake by leaving the answer at \[x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6},\dfrac{{3\pi }}{2}\], and write that there are three solutions of the given equation, which is wrong. We should always check whether these solutions satisfy the given equation or not to find which of those values are the possible solutions of the given equation.