Question

Find the general solution of $\cos x - \sin x = 1$.

Answer 1

Here, we need to simplify the equation and find a general solution of the equation. A general solution is a value of $x$ which is true for all numbers in a certain range. We will simplify the equation using algebraic and trigonometric identities, such that the left hand side and right hand side are the sine of some angle. Then, using the trigonometric equations, we will find the general solution of the equation.
Formula used: We will use the following formulas to solve the questions:

1. The square of the difference of two numbers is given by the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.
2. The sine of the double of an angle is twice the product of the sine and cosine of that angle, that is $\sin 2A = 2\sin A\cos A$.
3. Trigonometric equation: If $\sin x = 0$, then $x = n\pi$, where $n$ can be any integer.

Complete step by step solution:
We will use trigonometric identities to simplify the equation. Then, we will use trigonometric equations for sine to get the general solution of the given equation.
Squaring both sides, we get
$\begin{array}{l}{\left( {\cos x - \sin x} \right)^2} = {\left( 1 \right)^2}\\\ \Rightarrow {\left( {\cos x - \sin x} \right)^2} = 1\end{array}$
Substituting $a = \cos x$ and $b = \sin x$ in the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, we get
$\Rightarrow {\left( {\cos x - \sin x} \right)^2} = {\left( {\cos x} \right)^2} + {\left( {\sin x} \right)^2} - 2\left( {\cos x} \right)\left( {\sin x} \right)$
Rewriting the equation, we get
$\begin{array}{l} \Rightarrow {\left( {\cos x - \sin x} \right)^2} = {\cos ^2}x + {\sin ^2}x - 2\cos x\sin x \\\ \Rightarrow {\left( {\cos x - \sin x} \right)^2} = {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x\end{array}$
Now, the sum of the square of the sine and cosine of an angle is always equal to 1.
Therefore, we get
${\sin ^2}x + {\cos ^2}x = 1$
We know that $\sin 2A = 2\sin A\cos A$.
Therefore, substituting $A = x$ in the formula, we get
$\sin 2x = 2\sin x\cos x$
Substituting ${\sin ^2}x + {\cos ^2}x = 1$ and $2\sin x\cos x = \sin 2x$ in the equation ${\left( {\cos x - \sin x} \right)^2} = {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x$, we get
$\Rightarrow {\left( {\cos x - \sin x} \right)^2} = 1 - \sin 2x$
Substituting ${\left( {\cos x - \sin x} \right)^2} = 1$ in the equation, we get
$\Rightarrow 1 = 1 - \sin 2x$
Subtracting 1 from both sides of the equation, we get
$\begin{array}{l} \Rightarrow 1 - 1 = 1 - \sin 2x - 1\\\ \Rightarrow 0 = - \sin 2x\\\ \Rightarrow \sin 2x = 0\end{array}$
Now, we will use trigonometric equations for sine.
We know that if $\sin x = 0$, then $x = n\pi$, where $n$ can be any integer.
Therefore, since $\sin 2x = 0$, we get
$2x = n\pi ,{\rm{ }}n \in Z$
Dividing both sides by 2, we get
$\Rightarrow x = \dfrac{{n\pi }}{2},{\rm{ }}n \in Z$

Therefore, the general solution of the equation $\cos x - \sin x = 1$ is $x = \dfrac{{n\pi }}{2}$, where $n$ can be any integer.

Note:
We can also solve this question using trigonometric equations for cosine.
Dividing both sides of the equation $\cos x - \sin x = 1$ by $\sqrt 2$, we get
$\Rightarrow \dfrac{{\cos x - \sin x}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}$
Splitting the L.C.M. and rewriting the equation, we get
$\begin{array}{l} \Rightarrow \dfrac{{\cos x}}{{\sqrt 2 }} - \dfrac{{\sin x}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\\\ \Rightarrow \cos x \times \dfrac{1}{{\sqrt 2 }} - \sin x \times \dfrac{1}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\end{array}$
We know that the sine and cosine of the angle $\dfrac{\pi }{4}$ is $\dfrac{1}{{\sqrt 2 }}$.
Rewriting $\dfrac{1}{{\sqrt 2 }}$ as $\sin \dfrac{\pi }{4}$ and $\cos \dfrac{\pi }{4}$, we get
$\Rightarrow \cos x\cos \dfrac{\pi }{4} - \sin x\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4}$
The trigonometric identity for the cosine of the sum of two angles is given by the formula $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$.
Using the trigonometric identity in the equation, we get
$\Rightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = \cos \dfrac{\pi }{4}$
Now, we will use trigonometric equations for cosine.
We know that if $\cos x = \cos \theta$, then $x = 2n\pi \pm \theta$, where $n$ can be any integer and $\theta \in \left( {0,\pi } \right]$.
Therefore, since $\cos \left( {x + \dfrac{\pi }{4}} \right) = \cos \dfrac{\pi }{4}$, we get
$x + \dfrac{\pi }{4} = 2n\pi \pm \dfrac{\pi }{4}$
Subtracting $\dfrac{\pi }{4}$ from both sides, we get
$\Rightarrow x = 2n\pi \pm \dfrac{\pi }{4} - \dfrac{\pi }{4}$
$\Rightarrow x = 2n\pi + \dfrac{\pi }{4} - \dfrac{\pi }{4}$ or $x = 2n\pi - \dfrac{\pi }{4} - \dfrac{\pi }{4}$
Simplifying the equations, we get
$\Rightarrow x = 2n\pi$ or $x = 2n\pi - \dfrac{\pi }{2}$
Therefore, the general solution of the equation $\cos x - \sin x = 1$ is $x = 2n\pi$ or $x = 2n\pi - \dfrac{\pi }{2}$, where $n$ can be any integer.