I'm trying to solve an equation here but unfortunately I can't. The equation: $$ \cos x + \sin x = 0 $$ I'm trying to solve this by replacing $\cos x$ with $(1-t^2)/(1+t^2)$ and $\sin x$ with $2t/(1+t^2), t=\tan x/2, \ $ but I can't get the right solution. Also I have tried by squaring both sides but still nothing.
Can anyone help me ?
$\endgroup$ 18 Answers
$\begingroup$Note that $$\cos x + \sin x = 0 \iff \cos x = -\sin x$$
Now, $\cos x$ cannot equal zero, since if it did, $\sin x = -1$ or $\sin x = 1$, in which case the given equation isn't satisfied.
So we can divide by $\cos x$ to get $$1 = \dfrac{-\sin x}{\cos x} = -\tan x \iff \tan x = -1$$
Solving for $x$ gives us the values $x = \dfrac {3\pi}4 + k\pi$, where $k$ is any integer.
$\endgroup$ $\begingroup$Another way to solve this is to write $\cos x = (e^{ix}+e^{-ix})/2$ and $\sin x = (e^{ix}-e^{-ix})/2i$. The equation simplifies in a couple of easy steps to $e^{2ix}= e^{-\pi i/2}$. This is equivalent to $2x= -\pi /2 + 2\pi n$, so $x= -\pi /4 + \pi n$ for integral $n$.
$\endgroup$ 1 $\begingroup$Following where you got stuck and squaring both sides and you obtain
$$\sin^2 x+\cos^2x+2\sin x \cos x=0 \Rightarrow 1+\sin2x=0$$
Using $\sin 2x = 2\sin x \cos x$.
This means that
$$\sin 2x = -1$$ and hence $$2x = \frac {3\pi}{2}+2k\pi \Rightarrow x=\frac {3\pi}{4}+k\pi$$
$\endgroup$ $\begingroup$I'm going to go through this assuming that you're solving for solutions within $[0,2\pi]$.
$\cos x+\sin x=0$ $\implies \cos x=-\sin x$
With this, we can pull out our trusty old unit circle:
Then, we need to find any angles on the circle where $\cos x = -\sin x$
Sorry for the low res on the second image. But, as you can see, we have our angles. The solutions to $\sin x+\cos x=0$ between $[0,2\pi]$ are $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$.
Hope that helps!
$\endgroup$ 1 $\begingroup$Since $$\cos x+\sin x=\sqrt 2\sin(x+(\pi/4)),$$ you can solve $$\sin(x+(\pi/4))=0.$$
Hence, you'll have $$x+(\pi/4)=n\pi\ \ \ (n\in\mathbb Z).$$
$\endgroup$ $\begingroup$Hint:
$\cos x=-\sin x=\cos\left(\frac{1}{2}\pi+x\right)$
$\cos x=\cos\alpha$ gives $x=\pm\alpha+2k\pi$ for $k\in\mathbb{Z}$
$\endgroup$ $\begingroup$You already have some good answers, but just for the fun of it here's another way:
$$\cos x+\sin x=\cos x+\cos(π/2-x)=2\cos(π/4)\cos(x-π/4)=0,$$ which implies $$\cos(x-π/4)=0,$$ so that we have $$x-\fracπ4=\fracπ2+πk,$$ where $k$ is any integer. Finally this gives $$x=\frac{3π}{4}+πk,k\in\mathrm Z.$$
$\endgroup$ $\begingroup$$$\cos x + \sin x = 0$$
Multiply $\dfrac{\sqrt2}{2}$ to both sides:
$$\dfrac{\sqrt2}{2}\cos x + \dfrac{\sqrt2}{2}\sin x = 0$$
Or:
$$\cos\dfrac{\pi}{4}\cos x + \sin\dfrac{\pi}{4}\sin x = 0$$
Then:
$$\cos\left(x - \dfrac{\pi}{4}\right) = 0$$
So:
$$x - \dfrac{\pi}{4} = \dfrac{\pi}{2} + k\pi$$
Therefore:
$$x = \dfrac{3\pi}{4} + k\pi\qquad (k \in \Bbb Z)$$
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