Question

Solve the equation $\left( {\cos x - \sin x} \right)\left( {2\tan x + 2} \right) = 0$.

Answer 1

Here, we need to solve the given equation. First, we will rewrite the given equation and convert tangent to sine and cosine. Then, we will simplify the equation using algebraic identity for the product of sum and difference of two numbers to form two equations. Finally, we will use the general solution for an equation of the form $\tan x = \tan \alpha$ to find the possible values of $x$.

Formula Used:
We will use the following formulas:
The tangent of an angle $\theta$ can be written as $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.
The product of sum and difference of two numbers is given by the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.
If $\tan x = \tan \alpha$, then the general solution is given by $x = n\pi + \alpha$, where $n$ is an integer.

Complete step-by-step answer:
First, we will factor out 2 from the given equation, $\left( {\cos x - \sin x} \right)\left( {2\tan x + 2} \right) = 0$.
Thus, we get
$\Rightarrow 2\left( {\cos x - \sin x} \right)\left( {\tan x + 1} \right) = 0$
Dividing both sides of the equation by 2, we get
$\Rightarrow \left( {\cos x - \sin x} \right)\left( {\tan x + 1} \right) = 0$
Now, 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
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
Substituting $\tan x = \dfrac{{\sin x}}{{\cos x}}$ in the given equation $\left( {\cos x - \sin x} \right)\left( {\tan x + 1} \right) = 0$, we get
$\Rightarrow \left( {\cos x - \sin x} \right)\left( {\dfrac{{\sin x}}{{\cos x}} + 1} \right) = 0$
Taking the L.C.M. and adding the terms, we get
$\Rightarrow \left( {\cos x - \sin x} \right)\left( {\dfrac{{\sin x + \cos x}}{{\cos x}}} \right) = 0$
Rewriting the expression, we get
$\Rightarrow \dfrac{1}{{\cos x}}\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) = 0$
Multiplying both sides by $\cos x$, we get
$\Rightarrow \dfrac{{\cos x}}{{\cos x}}\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) = 0 \times \cos x$
Thus, we get
$\Rightarrow \left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) = 0$………..$\left( 1 \right)$
Substituting $a = \cos x$ and $b = \sin x$ in the algebraic identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$, we get
$\Rightarrow \left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) = {\left( {\cos x} \right)^2} - {\left( {\sin x} \right)^2}$
Therefore, we get
$\Rightarrow \left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) = {\cos ^2}x - {\sin ^2}x$……………….$\left( 2 \right)$
From the equations $\left( 2 \right)$ and $\left( 1 \right)$, we get
$\Rightarrow {\cos ^2}x - {\sin ^2}x = 0$
Adding ${\sin ^2}x$ to both sides of the equation, we get
$\begin{array}{l} \Rightarrow {\cos ^2}x - {\sin ^2}x + {\sin ^2}x = 0 + {\sin ^2}x\\\ \Rightarrow {\cos ^2}x = {\sin ^2}x\end{array}$
Dividing both sides by ${\cos ^2}x$, we get
$\begin{array}{l} \Rightarrow 1 = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\\\ \Rightarrow 1 = {\left( {\dfrac{{\sin x}}{{\cos x}}} \right)^2}\\\ \Rightarrow 1 = {\tan ^2}x\end{array}$
Taking the square root of both sides, we get
$\begin{array}{l} \Rightarrow \tan x = \pm \sqrt 1 \\\ \Rightarrow \tan x = \pm 1\end{array}$
Therefore, either $\tan x = 1$ or $\tan x = - 1$.
We know that the tangent of the angle measuring $\dfrac{\pi }{4}$ is 1.
Substituting $1 = \tan \dfrac{\pi }{4}$ in the equations, we get
$\Rightarrow \tan x = \tan \dfrac{\pi }{4}$ or $\tan x = - \tan \dfrac{\pi }{4}$
The expression of the form $- \tan x$ can be written as $\tan \left( { - x} \right)$.
Therefore, we get
$- \tan \dfrac{\pi }{4} = \tan \left( { - \dfrac{\pi }{4}} \right)$
Therefore, the equations become
$\Rightarrow \tan x = \tan \dfrac{\pi }{4}$ or $\tan x = \tan \left( { - \dfrac{\pi }{4}} \right)$
Now, we will find the general solution for the two equations.
If $\tan x = \tan \alpha$, then the general solution is given by $x = n\pi + \alpha$, where $n$ is an integer.
Since $\tan x = \tan \dfrac{\pi }{4}$, we get
$\Rightarrow x = n\pi + \dfrac{\pi }{4}$
Since $\tan x = \tan \left( { - \dfrac{\pi }{4}} \right)$, we get
$\begin{array}{l} \Rightarrow x = n\pi + \left( { - \dfrac{\pi }{4}} \right)\\\ \Rightarrow x = n\pi - \dfrac{\pi }{4}\end{array}$
Thus, we get the general solution of the equation $\left( {\cos x - \sin x} \right)\left( {2\tan x + 2} \right) = 0$ as $x = n\pi \pm \dfrac{\pi }{4}$, where $n$ is an integer.

Note: We can form the equations $\tan x = 1$ or $\tan x = - 1$ without converting tan to cosine and sine, and without using the algebraic identity.
We rewrote the given equation as $\left( {\cos x - \sin x} \right)\left( {\tan x + 1} \right) = 0$.
Thus, we get either $\left( {\cos x - \sin x} \right) = 0$ or $\left( {\tan x + 1} \right) = 0$.
Simplifying the equation $\left( {\cos x - \sin x} \right) = 0$, we get
$\begin{array}{l} \Rightarrow \cos x = \sin x\\\ \Rightarrow 1 = \tan x\end{array}$
Simplifying the equation $\left( {\tan x + 1} \right) = 0$, we get
$\Rightarrow \tan x = - 1$
Therefore, we have formed the equations $\tan x = 1$ or $\tan x = - 1$. The rest of the solution to find the general solution remains the same.