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

Mar 8, 2018

$x = 2 n \pi + \pi$ or $x = 2 n \pi - \frac{\pi}{2}$

#### Explanation:

Given: $\sin x + \cos x = - 1$

Square both sides

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

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

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

$2 \sin x \cos x = 0$

Either $\sin x = 0 \text{ or } \cos x = 0$

If $\sin x = 0$
$\cos x = - 1$

and $x = 2 n \pi + \pi$

If $\cos x = 0$
$\sin x = - 1$

and $x = 2 n \pi - \frac{\pi}{2}$

Mar 8, 2018

x= pi+2pin, n∈Z
x= (3pi)/2+2pin, n∈Z

#### Explanation:

1. Square both sides
${\left(\sin \left(x\right) + \cos \left(x\right)\right)}^{2} = {\left(- 1\right)}^{2}$
$\sin {\left(x\right)}^{2} + {\cos}^{2} x + 2 \sin x \cos x = 1$
$1 + 2 \sin x \cos x = 1$
$2 \sin x \cos x = 0$
$2 \sin x \left(\cos x\right) = 0$
$\sin x = 0$
$\cos x = 0$
x= pi+2pin, n∈Z
x= (3pi)/2+2pin, n∈Z
x=pi/2+2pin, n∈Z EXTRANEOUS
x=0+2pin, n∈Z EXTRANEOUS
Mar 8, 2018

$x = \frac{3}{2} \pi + 2 \pi n \text{ or } \pi + 2 \pi n$

#### Explanation:

$\sin \left(x\right) + \cos \left(x\right) = - 1$

$\implies \sin \left(x\right) = - 1 - \cos \left(x\right)$ equation-1

We have an identity

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

Use this to find the value of $\sin \left(x\right)$

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

$\implies \sin \left(x\right) = \pm \sqrt{1 - {\cos}^{2} \left(x\right)}$

We got two values for $\sin \left(x\right)$

$+ \sqrt{1 - {\cos}^{2} \left(x\right)} \text{and} - \sqrt{1 - {\cos}^{2} \left(x\right)}$

Put them one by one in equation-1.

$\implies + \sqrt{1 - {\cos}^{2} \left(x\right)} = - 1 - \cos \left(x\right)$

Squaring both sides

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

$\implies 0 = 2 {\cos}^{2} \left(x\right) + 2 \cos \left(x\right)$

Divide by two both sides

$\implies 0 = {\cos}^{2} \left(x\right) + \cos \left(x\right)$

$\implies 0 = \cos \left(x\right) \left(\cos \left(x\right) + 1\right)$

It gives $\cos \left(x\right) = 0$

We get $\sin \left(x\right) = - 1$

The solution for this is

$x = \frac{3}{2} \pi + 2 \pi n$

Here , $\pi = {180}^{\circ}$ and n is any integer.

Now , we also get $\cos \left(x\right) + 1 = 0$

$\implies \cos \left(x\right) = - 1$

It gives $\sin \left(x\right) = 0$ according to the given equation.

The solution for this is

$x = \pi + 2 \pi n$

If you put $\sin \left(x\right) = - \sqrt{1 - {\cos}^{2} \left(x\right)}$ you get the same results.

Hope it helps :)