# How do you solve sinx+cosx=1?

Jun 24, 2016

$x = 2 n \pi \mathmr{and} x = \left(4 n + 1\right) \frac{\pi}{2} , n = 0 , \pm 1 , \pm 2 , \pm 3. \ldots$

#### Explanation:

The given equation is equivalent to

$\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$.

This can be written as

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

The general solution of this equation ls

$x - \frac{\pi}{4} = 2 n \pi \pm \frac{\pi}{4} , n = 0 , \pm 1 , \pm 2 , \ldots$,

So, $x = 2 n \pi \mathmr{and} x = \left(4 n + 1\right) \frac{\pi}{2} , n = 0 , \pm 1 , \pm 2 , \pm 3. \ldots$

Having noted that there were 40K viewers for the answers by me,

Hero and Nghi, I think I could invoke more interest by including the

solutions for $\cos x - \sin x = 1$, and for that matter,

$\sec x \pm \tan x = 1$, that become

$\cos x - \sin x = 1$ and $\cos x + \sin x = 1$, upon multiplication by

cos x, when $x \ne$ an odd multiple of $\frac{\pi}{2}$.

For cos x - sin x = 1,

the general solution is

$x = 2 n \pi \mathmr{and} x = \left(4 n - 1\right) \frac{\pi}{2} , n = 0 , \pm 1 , \pm 2 , \pm 3. \ldots$

Note the change in the multiple from $\left(4 n + 1\right) \to \left(4 n - 1\right)$.

For $\sec x \pm \tan x = 1$, it is same sans $\left(4 n \pm 1\right) \frac{\pi}{2}$. It is just

$x = 2 n \pi$

See x-intercepts as graphical solutions.

Graph on uniform scale for solutions

$x = \ldots - 2 \pi , - \frac{3 \pi}{2} , 0 , \frac{\pi}{2} , 2 \pi . .$ of cos x + sin x = 1:
graph{y-cos x - sin x +1 = 0[-7 7 -3 4]}
Graph on uniform scale for solutions

$x = - 2 \pi , - \frac{\pi}{2} , 0 , \frac{3 \pi}{2} , 2 \pi , . .$ of $\cos x - \sin x = 1$:
graph{y-cos x + sin x +1 = 0[-7 7 -3 4]}

See combined graph for solutions $0 , \pm 2 \pi , \pm 4 \pi , \ldots$ of

$\sec x \pm \tan x = 1$:

graph{(y- sec x - tan x +1)(y- sec x+ tan x +1)=0[-13 13 -6.5 6.5]}

Feb 21, 2017

$x = \frac{\pi}{2} + 2 \pi n \mathmr{and} x = 2 \pi n$ where $n = 0 , \pm 1 , \pm 2 , \pm 3. . .$

#### Explanation:

If $\cos x + \sin x = 1$ then squaring both sides gives us:

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

Using the identities:
${\cos}^{2} x + {\sin}^{2} x = 1$ and $\sin 2 x = 2 \sin x \cos x$

The equation can be simplified to: $1 + \sin 2 x = 1$

Therefore $\sin 2 x = 0$

The values of $\theta$ at which $\sin \theta = 0$ are $\theta = n \pi$ where $n$ is an integer.

But here $\theta = 2 x$ so $x = n \frac{\pi}{2}$ for $n = 0 , \pm 1 , \pm 2 , \pm 3. . .$

However, since we squared the equation, we need to check all of these answers work in the original equation:

$\cos 0 + \sin 0 = 1 + 0 = 1$

$\cos \left(\frac{\pi}{2}\right) + \sin \left(\frac{\pi}{2}\right) = 0 + 1 = 1$

$\cos \pi + \sin \pi = - 1 + 0 = - 1$ This solution is not valid.

$\cos \left(3 \frac{\pi}{2}\right) + \sin \left(3 \frac{\pi}{2}\right) = 0 - 1 = - 1$ This solution is not valid either.

When we get to $2 \pi$ the graphs repeat, so $2 \pi , 5 \frac{\pi}{2} , 4 \pi , 9 \frac{\pi}{2.} . .$ are all valid, but the other solutions aren't.

This can be written as:
$x = \frac{\pi}{2} + 2 \pi n \mathmr{and} x = 2 \pi n$ where $n = 0 , \pm 1 , \pm 2 , \pm 3. . .$

Feb 24, 2017

$x = \left(2 k + 1\right) \left(\frac{\pi}{2}\right)$
$x = 2 k \pi$

#### Explanation:

Use trig identity:
$\sin x + \cos x = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right)$
We get:
$\sqrt{2} \cos \left(x - \frac{\pi}{4}\right) = 1$ --> $\cos \left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Trig table and unit circle give -->
$\left(x - \frac{\pi}{4}\right) = \pm \frac{\pi}{4} + 2 k \pi$--> 2 solutions -->
$a . x - \frac{\pi}{4} = \frac{\pi}{4} + 2 k \pi$ -->
$x = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} + 2 k \pi$, and
$b . x - \frac{\pi}{4} = - \frac{\pi}{4} + 2 k \pi$
$x = \frac{\pi}{4} - \frac{\pi}{4} = 0 + 2 k \pi = 2 k \pi$