# Solve the Following Equation: Sin X + Cos X = 1 - Mathematics

Sum

Solve the following equation:

$\sin x + \cos x = 1$

#### Solution

Given:
$\sin x + \cos x = 1$      ...(i)

The equation is of the form
$a \sin \theta + b \cos \theta = c$, where
$a = 1, b = 1$ and $c = 1$.
Let: $a = r \sin \alpha$ and $b = r \cos \alpha$
Now,
$r = \sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$ and

$\tan\alpha = \frac{b}{a} = 1 \Rightarrow \alpha = \frac{\pi}{4}$

On putting

$a = 1 = r \sin \alpha$ and $b = 1 = r \cos \alpha$ in equation (i), we get:
$r \sin \alpha \sin x + r \cos \alpha \cos x = 1$

$\Rightarrow r \cos ( x - \alpha) = 1$

$\Rightarrow \sqrt{2} \cos \left( x - \frac{\pi}{4} \right) = 1$

$\Rightarrow \cos \left( x - \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$

$\Rightarrow \cos \left( x - \frac{\pi}{4} \right) = \cos \frac{\pi}{4}$

$\Rightarrow x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}, n \in Z$

On taking positive sign, we get:
$x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{4}$
$\Rightarrow x = 2n\pi + \frac{\pi}{4} + \frac{\pi}{4}$
$\Rightarrow x = 2n\pi + \frac{\pi}{2}, n \in Z$
On taking negative sign, we get:
$x - \frac{\pi}{4} = 2m\pi - \frac{\pi}{4}$
$\Rightarrow x = 2m\pi, m \in Z$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 6.3 | Page 22