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?

Answer 1

Hint : To solve the above given question we have to know the trigonometric functions and also should know about intervals, some basic formulas of trigonometric functions and about factorisation.
Here we need to find a number of solutions for the given question.

Complete step-by-step answer :
Let us consider the given question $\tan x + \sec x = 2\cos x$
$\Rightarrow$ $\dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}} = 2\cos x$ [here $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$ ]
$\Rightarrow$ $\dfrac{{\sin x + 1}}{{\cos x}} = 2\cos x$ [ the denominator is same therefore LCM of denominator remains same]
$\Rightarrow$ $1 + \sin x = 2{\cos ^2}x$ [ $\cos x$ is multiplied to right and side]
$\Rightarrow$ $1 + \sin x = 2(1 - {\sin ^2}x)$ [ here ${\cos ^2}x = (1 - {\sin ^2}x)$ ]
$\Rightarrow$ $1 + \sin x = 2 - 2{\sin ^2}x$
$\Rightarrow$ $2{\sin ^2}x + \sin x + 1 - 2 = 0$ [ rearranging the terms and making right hand side equal to zero]
$\Rightarrow$ $2{\sin ^2}x + \sin x - 1 = 0$ [ applying factorisation to find the factors]
$\Rightarrow$ $2{\sin ^2}x + 2\sin x - \sin x - 1 = 0$
$\Rightarrow$ $2\sin x(\sin x + 1) - 1(\sin x + 1) = 0$ [taking common factors and equating to zero]
$\Rightarrow$ $(2\sin x - 1)(\sin x + 1) = 0$
$\Rightarrow$ $\sin x = \dfrac{1}{2}$ or $\sin x = - 1$ [equating the terms to zero]
$\Rightarrow$ $x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$ or $x = \dfrac{{3\pi }}{2}$
[ here we have considered $x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$ ]
$\because$ $x = \dfrac{{3\pi }}{2}$ is not possible as original equation have $\tan \theta$ and $\tan \left( {\dfrac{{3\pi }}{2}} \right) = - \infty$
$\therefore$ $x = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$
So, two solutions for $x$ are possible.
Therefore, we can say that for the given question $\tan x + \sec x = 2\cos x$ there are two solutions.
So, the correct answer is “2 solutions”.

Note : Trigonometric functions say there are real functions which relate an angle of a right-angled triangle to ratios of two side lengths.
Tangent and cotangent functions are positive in the third quadrant.
Factorisation says factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind and also factorisation is unique up to the order of the factors.
$\left[ {0,2\pi } \right]$ is in the fourth quadrant in which all the six functions such as sin, cos, tan, cosine, sec, and cot functions are positive.