Question

Find the number of roots of the equation $\tan x + \sec x = 2\cos x$ in the interval $\left[ {0,2\pi } \right]$
A. 1
B. 2
C. 3
D. 4

Answer 1

We convert all the trigonometric functions into simpler forms of sine and cosine. Take LCM and form a quadratic equation. Use the formula of trigonometry ${\cos ^2}x = 1 - {\sin ^2}x$ to convert the complete equation in terms of sine. Find the roots of the equation by factorization method.

• $\sec x = \dfrac{1}{{\cos x}};\tan x = \dfrac{{\sin x}}{{\cos x}}$
• Roots of an equation $a{x^n} + b{x^{n - 1}} + ........c = 0$ are those values of x which give the value of the equation equal to zero when substituted in the equation. Let y be a root of the equation, we can write $(x - y) = 0$ is a factor of the equation. On equating the factor we write $x = y$ is a root of the equation.
• Factorization method: If ‘p’ and ‘q’ are the roots of a quadratic equation, then we can say the quadratic equation is $(x - p)(x - q) = 0$

Complete step-by-step answer:
We are given the equation $\tan x + \sec x = 2\cos x$............… (1)
Substitute the value of $\sec x = \dfrac{1}{{\cos x}};\tan x = \dfrac{{\sin x}}{{\cos x}}$ in equation (1)
$\Rightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}} = 2\cos x$
Take LCM of the fractions on LHS of the equation
$\Rightarrow \dfrac{{\sin x + 1}}{{\cos x}} = 2\cos x$
Multiply both sides of the equation by $\cos x$
$\Rightarrow \dfrac{{\sin x + 1}}{{\cos x}} \times \cos x = 2\cos x \times \cos x$
Cancel same terms from numerator and denominator in LHS of the equation
$\Rightarrow \sin x + 1 = 2{\cos ^2}x$
Substitute the value of ${\cos ^2}x = 1 - {\sin ^2}x$ in RHS of the equation
$\Rightarrow \sin x + 1 = 2(1 - {\sin ^2}x)$
$\Rightarrow \sin x + 1 = 2 - 2{\sin ^2}x$
Shift all terms to LHS of the equation
$\Rightarrow \sin x + 1 - 2 + 2{\sin ^2}x = 0$
$\Rightarrow 2{\sin ^2}x + \sin x - 1 = 0$..............… (2)
Now we factorize the terms of equation (2)
We can write $\sin x = 2\sin x - \sin x$
Substitute the value of $\sin x = 2\sin x - \sin x$ in equation (2)
$\Rightarrow 2{\sin ^2}x + 2\sin x - \sin x - 1 = 0$
Take $2\sin x$common from first two terms and -1 common from last two terms of the equation
$\Rightarrow 2\sin x(\sin x + 1) - 1(\sin x + 1) = 0$
Collect the factors
$\Rightarrow (\sin x + 1)(2\sin x - 1) = 0$
Since we know the roots of an equation are found by equating the factors to zero.
Put $(\sin x + 1) = 0$
$\Rightarrow \sin x = - 1$
We know sine has value -1 when the angle is $\dfrac{{3\pi }}{2}$
$\Rightarrow x = \dfrac{{3\pi }}{2}$
Since we cannot take the value of $\cos x \ne 0$, so we will neglect this value of x as $\cos \dfrac{{3\pi }}{2} = 0$ which will make the initial equation zero.
Also, $(2\sin x - 1) = 0$
$\Rightarrow 2\sin x = 1$
Divide both sides by 2
$\Rightarrow \sin x = \dfrac{1}{2}$
We know sine has value $\dfrac{1}{2}$ when the angle is $\dfrac{\pi }{6},\dfrac{{5\pi }}{6}$
$\Rightarrow x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$
So, $\dfrac{\pi }{6},\dfrac{{5\pi }}{6}$ are two roots of the equation $\tan x + \sec x = 2\cos x$
$\therefore$Number of roots is 2.

$\therefore$Correct option is B.

Note: Students are likely to make the mistake of writing that there are three roots of the equation as they don’t check for the initial condition. Also, many students assume the value of roots will be $-1$ and $\dfrac{1}{2}$ but those are the values of the function $\sin x$ obtained from solving the quadratic equation.