Question

How do you solve \[\tan x - 2 - 3\cot x = 0\]?

Answer 1

Hint: We transform the cotangent value to tangent value and form the quadratic equation in terms of tangent. Assume tangent of x as a variable as solve for the variable using factorization method. Substitute back the value of the variable and take inverse tangent function to calculate the value of x.
* \[\cot \theta = \dfrac{1}{{\tan \theta }}\]
* 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 \[\tan x - 2 - 3\cot x = 0\]
Substitute the value of \[\cot x = \dfrac{1}{{\tan x}}\]in the equation
\[ \Rightarrow \tan x - 2 - \dfrac{3}{{\tan x}} = 0\]
Take LCM on left hand side of the equation
\[ \Rightarrow \dfrac{{{{\tan }^2}x - 2\tan x - 3}}{{\tan x}} = 0\]
Cross multiply the value from denominator of left hand side of the equation to right hand side of the equation
\[ \Rightarrow {\tan ^2}x - 2\tan x - 3 = 0\]
Now we can see this is a quadratic equation in terms of tangent of x
Substitute \[\tan x = y\]
\[ \Rightarrow {y^2} - 2y - 3 = 0\]
We can write the equation such that the coefficient of x is broken in such a way that its product equals product of coefficient of other two terms and sum equals coefficient of x.
\[ \Rightarrow {y^2} + y - 3y - 3 = 0\]
Take y common from first two terms and -3 common from last two terms
\[ \Rightarrow y(y + 1) - 3(y + 1) = 0\]
Collect the factors
\[ \Rightarrow (y + 1)(y - 3) = 0\]
Equate the factors to 0
\[ \Rightarrow y + 1 = 0\] and \[y - 3 = 0\]
Shift constant values to right hand side
\[ \Rightarrow y = - 1\] and \[y = 3\]
Now substitute the value of y back i.e. put \[y = \tan x\]
\[ \Rightarrow \tan x = - 1\] and \[\tan x = 3\]
Take inverse trigonometric function on both sides of the equation
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan x} \right) = {\tan ^{ - 1}}( - 1)\] and \[{\tan ^{ - 1}}\left( {\tan x} \right) = {\tan ^{ - 1}}(3)\]
Cancel tangent from inverse tangent function
\[ \Rightarrow x = {\tan ^{ - 1}}( - 1)\] and \[x = {\tan ^{ - 1}}(3)\]

\[\therefore \]Solution of the equation \[\tan x - 2 - 3\cot x = 0\] are \[x = {\tan ^{ - 1}}( - 1)\]and \[x = {\tan ^{ - 1}}(3)\].

Note:
 Many students get confused while solving for the value of variables in the equation and write the value of \[\tan x\] as the answer because the equation is formed in \[\tan x\]. Keep in mind the variable here is x, so we will calculate the value of x.