Finding the Zeroes: $f(x)=\sin(x)+\cos(x)$ I am a bit clueless on how to find the zeroes of this function. I know that the function would be zero at $x=-\frac{5\pi}4,-\frac{\pi}4,\frac{3\pi}4$, and etc by graphing the function.  How would I find the zeroes of this function without graphing?

$$f(x)=\sin(x)+\cos(x)$$

 A: $$
\sin x+\cos x=0\iff \sin x=-\cos x.
$$
If $\cos x=0$, then $\sin x\not=0$, so the equation doesn't hold. Thus, the above holds iff
$$
\tan x=-1.
$$
This occurs iff we have a $45$ degree angle in the second or fourth quadrant. I.e., iff
$$
x=\frac{3\pi}{4}+\pi n
$$
where $n\in\mathbb{Z}$.
A: Hint:
$$\sin(x)+\cos(x)=\sqrt 2 \left( \frac{1}{\sqrt{2}}\sin(x)+\frac{1}{\sqrt{2}}\cos(x)\right)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$
A: Notice that according to the sum of angles formulas,
$$\sin(x)+\cos(x)=\sqrt2\sin\left(x+\frac\pi4\right)$$
Setting this equal to $0$, we now get
$$\sin\left(x+\frac\pi4\right)=0$$
$$x=...$$
A: Another method that has some generalization, as it works for any pair of shifted functions:
$\sin(x)$ and $\cos(x)$ are shifts of each other, which means that there exists a $k$ such that $\sin(x+k)=\cos(x)$ (in our case, $k=\pi/2$. Rewriting the equation using that, we can obtain the expression $$f(x)=\sin\left(x-\frac{k}{2}\right)+\sin\left(x+\frac{k}{2}\right)$$
to make $f(x)=0$, we just need to find the appropriate value of $x$ to satisfy this equation. In the case of $\sin(x)$, it's rapidly apparent that the answer is $\pi n + 3\pi/4$ because that gives us
$$\sin\left(\pi n + \frac{1}{2}\right)+\sin\left(\pi(n+1)\right)=\pm(-1+1)=0$$ and a few quick arguments concerning derivatives suffice to show that there aren't intermediate zeroes.
A: Zeros occur when $f(x)=0$, so we we need values such that $\cos(x)=-\sin(x)$. Dividing both sides by $\cos(x)$, we have $\tan(x)=-1$. The solutions are precisely those you list.
