# Solving $\cos x+\sin x-1=0$

How does one solve this equation?

$$\cos {x}+\sin {x}-1=0$$

I have no idea how to start it.

Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?

Rewrite as:

$$\cos(x)+\sin(x)=1$$ Then square: $$\cos^2(x)+2\cos(x)\sin(x)+\sin^2(x)=1$$ Note an important identity: $$1+2\cos(x)\sin(x)=1$$ Then simplify and note another identity: $$\sin(2x)=0$$

Can you take it from here?

• Someone downvoted every answer in this question. – user223391 Jun 1 '15 at 0:16
• Down voters need to be neutralized. +1 for this solid answer. – Mark Viola Jun 1 '15 at 1:06
• fwiw you can skip the last identify and go for $\cos x = 0$ or $\sin x = 0$. – djechlin Oct 19 '16 at 1:51

Given $$\color{blue}{\cos x+\sin x-1=0}$$$$\implies \cos x+\sin x=1$$ Divide both sides by $\color{blue}{\sqrt{2}}$ we get $$\frac{1}{\sqrt{2}}\cos x+ \frac{1}{\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$$ $$\cos x\cos\frac{\pi}{4}+\sin x\sin\frac{\pi}{4}=\cos\frac{\pi}{4}$$ Using formula $\color{purple}{\cos A\cos B+\sin A\sin B=\cos(A-B)}$, we get $$\color{green}{\cos\left(x-\frac{\pi}{4}\right)=\cos\frac{\pi}{4}}$$ As there is no information about the unknown value $x$ hence writing the general solutions as follows $$x-\frac{\pi}{4}=2n\pi\pm \frac{\pi}{4}$$$$\implies x=2n\pi\pm \frac{\pi}{4}+\frac{\pi}{4}$$$$\implies \color{blue}{x=2n\pi} \quad \text{&}\quad \color{blue}{x=2n\pi+\frac{\pi}{2}}$$ Where, $\color{blue}{n \space \text{is any integer}}$

As $\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(\cos x+\sin x\right)$ by the angle addition formula we find that: \begin{aligned} \cos x+\sin x-1&=0\\ \implies\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)&=1\\ \implies\sin\left(x+\frac{\pi}{4}\right)&=\frac{\sqrt{2}}{2}\\ \implies x+\frac{\pi}{4}&=\frac{\pi}{4}+2\pi n,\frac{3\pi}{4}+2\pi n\\ \implies x&=2\pi n,\frac{\pi}{2}+2\pi n \end{aligned} for $n\in\mathbb{Z}$.

$\cos x+\sin x=\sqrt2\cos\Bigl(x-\dfrac\pi4\Bigr)$, hence the equation is equivalent to: $$\cos\Bigl(x-\frac\pi4\Bigr)=\frac1{\sqrt2}\iff x-\frac\pi4\equiv\pm\frac\pi4\mod 2\pi\iff x\equiv 0,\,\frac\pi2\mod2\pi.$$

you can use: $$\cos(x)= \sqrt{1-\sin(x)^{2}}$$ $$\text{then let}\qquad \sin(x)=t\\ \sqrt{1-t^{2}}+t-1=0$$ now you can continue

• And what about the cases where $\cos x \neq \sqrt{1-\sin^2 x}$ ? – user228113 May 31 '15 at 22:44
• $$in \;this \;case \;: cos(x)=-\sqrt{1-sin(x)}\leq 0\\but :cos(x)=1-sin(x)\leq 0\; so \;cos(x)=0\\in \; this \; case \; \sqrt{1-sin(x)}=-\sqrt{1-sin(x)}=0$$ – mounir ben salem May 31 '15 at 22:52
• Did you know that your answer can contain text that is no between \$...\$? – Mike Pierce May 31 '15 at 22:55
• yes i knew that $_$ – mounir ben salem May 31 '15 at 22:59

I'll throw my hat in the ring to get a picture in edgewise. :-)

I'll use $\theta$ for the angle, rather than $x$ (to avoid confusion with the $x$-coordinate). If we then let $x = \cos \theta, y = \sin \theta$, then the allowable results are along the red circle with $x^2+y^2 = 1$ (since $\cos^2 \theta + \sin^2 \theta = 1$). The solutions of the given equation are at the intersections of the blue line $x+y = 1$ with that red circle, yielding $(\cos \theta, \sin \theta) = (1, 0)$ and $(0, 1)$.

This in turn yields

$$\theta = 2k\pi, \qquad \theta = 2k\pi + \frac{\pi}{2}$$

for $k$ an integer.

By inspection, it is obvious, that: $1-\sin{x} \equiv (\cos{\frac{x}{2}} - \sin{\frac{x}{2}})^2$.

From the half angle expansions, $\cos{x} \equiv (\cos{\frac{x}{2}} - \sin{\frac{x}{2}})(\cos{\frac{x}{2}} + \sin{\frac{x}{2}})$.

The equation is thus equivalent to:

$(\cos{\frac{x}{2}} - \sin{\frac{x}{2}})(2\sin{\frac{x}{2}}) = 0$

Which can then be solved using standard methods.

starting from, $$\cos x+\sin x-1=0$$ $$(\sin x+\cos x)^2=1^2$$ completing square, i get $$\sin^2 x+\cos^2 x+2\sin \cos x=1$$ $$1+\sin 2x=1$$$$\sin 2x=0$$ $$2x=k\pi$$ $$x=\frac{k\pi}{2}$$ where, $k=0, \pm 1, \pm2, \ldots$

• $\equiv$ Zachary Selk's answer? – snulty Oct 19 '16 at 1:11
• $x=k\pi/2$ is not a solution to $\cos x+\sin x-1=0$ when $k=2$. Or $3$. – Barry Cipra Oct 19 '16 at 1:47