# How do you solve sinx+cosx=1?

Dec 28, 2016

The answer is $S = \left\{2 k \pi , \frac{\pi}{2} + 2 k \pi\right\}$, $k \in \mathbb{Z}$

#### Explanation:

We need

$\sin \left(A + B\right) = \sin A \cos B + \sin B \cos A$

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

We compare this equation to

$r \sin \left(x + a\right) = 1$

$r \sin x \cos a + r \cos x \sin a = 1$

$\sin x + \cos x = 1$

Therefore,

$r \cos a = 1$ and $r \sin a = 1$

So,

${\cos}^{2} a + {\sin}^{2} a = \frac{1}{r} ^ 2 + \frac{1}{r} ^ 2 = \frac{2}{r} ^ 2 = 1$

${r}^{2} = 2$, $\implies ,$r=sqrt2#

and $\tan a = 1$, $\implies$, $a = \frac{\pi}{4}$

Therefore,

$\sqrt{2} \sin \left(x + \frac{\pi}{4}\right) = 1$

$\sin \left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

$x + \frac{\pi}{4} = \frac{\pi}{4} + 2 k \pi$, $\implies$, $x = 2 k \pi$

and

$x + \frac{\pi}{4} = 3 \frac{\pi}{4} + 2 k \pi$, $\implies$, $x = \frac{\pi}{2} + 2 k \pi$

The solutions are $S = \left\{2 k \pi , \frac{\pi}{2} + 2 k \pi\right\}$, $k \in \mathbb{Z}$

Dec 28, 2016

Here's an alternative answer. Square both sides.

${\left(\sin x + \cos x\right)}^{2} = {1}^{2}$

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

Use the identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$.

$1 + 2 \sin x \cos x = 1$

$2 \sin x \cos x = 0$

Use the identity $2 \sin \theta \cos \theta = \sin 2 \theta$:

$\sin 2 x = 0$

$2 x = 0 , \pi$

$x = 2 \pi n , \frac{\pi}{2} + 2 \pi n$, where $n$ is an integer.

Hopefully this helps!