# How do you solve sin x + cos x = 1?

Refer to explanation

#### Explanation:

Squaring both sides of the equation yields to

$\sin x + \cos x = 1 \implies {\left(\sin x + \cos x\right)}^{2} = {1}^{2} \implies {\sin}^{2} x + {\cos}^{2} x + 2 \cos x \sin x = {\sin}^{2} x + {\cos}^{2} x \implies \sin x \cdot \cos x = 0 \implies \sin x = 0 \mathmr{and} \cos x = 0$

The solutions to $\sin x = 0 \mathmr{and} \cos x = 0$ are $0 , 90 , 270 , 360$ but $270$ does not satisfy the original equation.
So the solutions are ${0}^{o} , {90}^{o} , {360}^{o}$

The following identities were used

${\sin}^{2} x + {\cos}^{2} x = 1$

${\left(a + b\right)}^{2} = {a}^{2} + {b}^{2} + 2 a b$

Sep 23, 2015

Solve sin x + cos x = 1

Ans: x = 0 and $x = \frac{\pi}{2}$ and $x = 2 \pi$

#### Explanation:

Use the trig identity: $\sin x + \cos x = \sqrt{2.} \sin \left(x + \frac{\pi}{4}\right)$
We get:
$\sin \left(x + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ --> $\left(x + \frac{\pi}{4}\right) = \sin \frac{\pi}{4}$
and $\left(x + \frac{\pi}{4}\right) = \sin \left(\frac{3 \pi}{4}\right)$

a. $x + \frac{\pi}{4} = \frac{\pi}{4}$ --> $x = 0$ and $x = 2 \pi$
b. $x + \frac{\pi}{4} = \left(\frac{3 \pi}{4}\right)$ --> $x = \left(\frac{3 \pi}{4}\right) - \frac{\pi}{4} = \frac{\pi}{2}$

Check.
x = 0 --> sin x = 0; cos x = 1. Then 0 + 1 = 1. OK
$x = \frac{\pi}{2}$ --> sin x = 1; cos x = 0. Then 1 + 0 = 1 . OK