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

Question

How do we solve $\sin x+\cos x=1$?

I have solved it easily by making the substitution, $\sin x+\cos x=1=R\sin(x+a)$ which gives the solutions $x=2n\pi, 2n\pi+\pi/2$. But when I do as follows $$ (\sin x+\cos x)^2=1\implies 2\sin x\cos x=0\implies \sin 2x=0\implies2x=n\pi $$ For even and odd $$ 2x=2m\pi\text{ or }2x=(2m+1)\pi\\x=m\pi\text{ or } x=m\pi+\frac{\pi}{2} $$ What am I missing here ?

Consider the last part of Lab Bhattacharjee's answer below and reduce your equation to $$a.sin(x)+b.cos(x)=c$$ with $a^2+b^2=1$ by setting $$a:=\frac{A}{\sqrt{A^2+B^2}}\ ;\ B:=\frac{B}{\sqrt{A^2+B^2}}\ ;\ c:=\frac{C}{\sqrt{A^2+B^2}}$$ Then set $$a:=sin(\alpha)\ ;\ b=cos(\alpha)$$ you will be on the right track ! — Duchamp Gérard H. E., Aug 12 '17 at 12:26

score 5 · Accepted Answer · answered Aug 12 '17 at 06:02

5

You squared your equation. That gives you extra solutions. Specifically, your second approach picks up all solutions to $\sin x+\cos x=-1$ as well.

answered Aug 12 '17 at 06:02

Arthur

199,419

thnx. I make this mistake all the time by squaring, though not a mistake to square. – Sooraj S Aug 12 '17 at 06:11
2

It's not a mistake to square, that's right. You just need to remember that you have, and check solutions. – Arthur Aug 12 '17 at 06:25

score 1 · Answer 2 · answered Aug 12 '17 at 06:02

1

This method gives the solution to $\sin x+\cos x=1$ or $\sin x+\cos x=-1$.

Both the equations becomes $(\sin x+ \cos x) ^2=1$ after squaring.

answered Aug 12 '17 at 06:02

CY Aries

23,393

Michael Rozenberg · Answer 3 · 2017-08-12T06:22:49.550

$x=1$ is not equal to $x^2=1$.

For example: $-1=1$ is wrong, but $(-1)^2=1^2$ is true.

I think it's better to solve your equation by the following way.

We need to solve $$\frac{1}{\sqrt2}\sin{x}+\frac{1}{\sqrt2}\cos{x}=\frac{1}{\sqrt2}$$ or $$\sin(45^{\circ}+x)=\sin45^{\circ},$$ which gives $$45^{\circ}+x=45^{\circ}+360^{\circ}k,$$ which is $x=360^{\circ}k$, $k\in\mathbb Z$ or $$45^{\circ}+x=135^{\circ}+360^{\circ}k,$$ which is $x=90^{\circ}+360^{\circ}k$, $k\in\mathbb Z$ and we got the answer: $$\{90^{\circ}+360^{\circ}k,360^{\circ}k|k\in\mathbb Z\}$$

score 1 · Answer 4 · answered Aug 12 '17 at 06:32

One way is this: if $\sin x + \cos x = 1$, then both $\sin x$ and $\cos x$ are non negative, otherwise the other one would be greater than $1$. If one of them is different from $0$ or $1$, one has $$1 = \sin^2 x + \cos^2 x \lt \sin x + \cos x = 1$$ This is impossible. The only remaining possibilities are $(\sin x, \cos x) = (0, 1) $ or $(1, 0)$, hence $x = 2 k \pi$ or $\frac{\pi}{2} + 2 k \pi$.

score 1 · Answer 5 · answered Aug 12 '17 at 06:37

1

HINT: use that $$\sin(x)+\cos(x)=\sqrt{2} \sin \left(x+\frac{\pi }{4}\right)$$

answered Aug 12 '17 at 06:37

Dr. Sonnhard Graubner

95,283

score 1 · Answer 6 · answered Aug 12 '17 at 09:32

1

HINT:

Avoid squaring as it generally introduces extraneous root(s).

Here use double angle formula $\sin2A=2\sin A\cos A$ and $\cos2A=12-\sin^2A$

OR in general cases $$A\sin x+B\cos x=C$$ use Weierstrass Substitution

answered Aug 12 '17 at 09:32

lab bhattacharjee

274,582

score 0 · Answer 7 · answered Aug 12 '17 at 07:46

0

You can divide both sides by $\sqrt 2$ to get $$\cos (\frac {π}{4}-x) =1$$ which can be further be solved easily.

answered Aug 12 '17 at 07:46

Rohan Shinde

9,737

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

7 Answers7