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 $\cos{x}+\sin{x}$?
Thanks in advance!
$\endgroup$ 48 Answers
$\begingroup$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?
$\endgroup$ 3 $\begingroup$Given $$\color{blue}{\cos x+\sin x-1=0} $$$$\cos x+\sin x=1 $$ 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 $$x-\frac{\pi}{4}=2n\pi\pm \frac{\pi}{4}$$$$x=2n\pi\pm \frac{\pi}{4}+\frac{\pi}{4}$$$$ \color{}{x=2n\pi} \quad \text{Or}\quad \color{}{x=2n\pi+\frac{\pi}{2}} $$ $$\color{blue}{x\in\{2n\pi\}\cup\{2n\pi+\frac{\pi}{2}\}}$$ Where, $\color{}{n \space \text{is any integer}}$ i.e. $\ n=0, \pm1, \pm2,\pm3, \ldots$
$\endgroup$ 0 $\begingroup$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{equation} \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} \end{equation} for $n\in\mathbb{Z}$.
$\endgroup$ 0 $\begingroup$$\cos x+\sin x=\sqrt2\cos\Bigl(x-\dfrac\pi4\Bigr)$, hence the equation is equivalent to: $$\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.$$
$\endgroup$ 0 $\begingroup$you can use: $$ \cos(x)= \sqrt{1-\sin(x)^{2}}$$ $$ \text{then let}\qquad \sin(x)=t\\ \sqrt{1-t^{2}}+t-1=0$$ now you can continue
$\endgroup$ 4 $\begingroup$I'll throw my hat in the ring to get a picture in edgewise. :-)
I'll use $\theta$ for the angle, rather than $x$ (to avoid confusion with the $x$-coordinate). If we then let $x = \cos \theta, y = \sin \theta$, then the allowable results are along the red circle with $x^2+y^2 = 1$ (since $\cos^2 \theta + \sin^2 \theta = 1$). The solutions of the given equation are at the intersections of the blue line $x+y = 1$ with that red circle, yielding $(\cos \theta, \sin \theta) = (1, 0)$ and $(0, 1)$.
This in turn yields
$$ \theta = 2k\pi, \qquad \theta = 2k\pi + \frac{\pi}{2} $$
for $k$ an integer.
$\endgroup$ $\begingroup$By inspection, it is obvious, that: $1-\sin{x} \equiv (\cos{\frac{x}{2}} - \sin{\frac{x}{2}})^2$.
From the half angle expansions, $\cos{x} \equiv (\cos{\frac{x}{2}} - \sin{\frac{x}{2}})(\cos{\frac{x}{2}} + \sin{\frac{x}{2}})$.
The equation is thus equivalent to:
$(\cos{\frac{x}{2}} - \sin{\frac{x}{2}})(2\sin{\frac{x}{2}}) = 0$
Which can then be solved using standard methods.
$\endgroup$ $\begingroup$starting from, $$\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, $k=0, \pm 1, \pm2, \ldots$
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