Point of intersection of $f(x)=\sin(2x)+\cos(2x)$ and the $x$-axis How can I algebraically (without looking at the graph) find the point of intersection of $f(x)=\sin(2x)+\cos(2x)$ and $x$-axis, in the interval $[0, \pi]$?
 A: Hint:
You have to solve the equation
$$
\sin 2x + \cos 2 x=0
$$
If $\cos  2x \ne 0$, dividing by $\cos 2 x$ you find 
$$
\tan 2x =-1
$$
You can solve this ?
Now test if $\cos 2x=0$ can gives another solution (obviously not).
A: Hint
$$\sin(2x)+\cos(2x)=\frac 2 {\sqrt 2} \Big(\frac  {\sqrt 2} 2 \sin(2x)+\frac {\sqrt 2} 2 \cos(2x)\Big)=\frac 2 {\sqrt 2} \sin(2x+\frac \pi 4)=\frac 2 {\sqrt 2} \cos(2x-\frac \pi 4)$$
Chose the form you prefer among the last expressions.
A: Notice, $$f(x)=\sin 2x+\cos 2x$$ $$\implies f(x)=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos 2x+\frac{1}{\sqrt{2}}\sin 2x\right)$$ $$=\sqrt{2}\left(\cos 2x\cos\frac{\pi}{4}+\sin 2x\sin\frac{\pi}{4}\right)$$ $$=\sqrt{2}\cos \left(2x-\frac{\pi}{4}\right)$$ Now, for the intersection of $f(x)$ with the x-axis, we have $f(x)=0$ $$\implies \sqrt{2}\cos \left(2x-\frac{\pi}{4}\right)=0 $$ $$\implies \cos \left(2x-\frac{\pi}{4}\right)=0 $$ Writing the general solution as follows $$\implies 2x-\frac{\pi}{4}=\frac{(2n+1)\pi}{2}$$ $$\implies x=\frac{(2n+1)\pi}{4}+\frac{\pi}{8}$$ $$\implies x=\frac{(4n+3)\pi}{8}$$ Where, $n$ is any integer 
For given interval $[0, \pi]$, put $n=0$ & $n=1$ in the general solution, we get  $$\color{blue}{x\in \left\{\frac{3\pi}{8}, \frac{7\pi}{8}\right\}}$$
A: Hint: We have $$\bbox[border: blue 1px solid, 10px]{\sin 2x + \cos 2x \equiv \sqrt{2} \sin \left(2x + \frac{\pi}{4}\right).}$$
This means you need only solve $$\sin \left(2x + \frac{\pi}{4}\right) = 0$$
We know that $\arcsin 0 = 0$, can you continue from there?
A: Hint: You want $\sin 2x = - \cos 2x$. What does this tell you about $\tan 2x$?
A: Since $\cos(2x)$ and $\sin(2x)$ can't simultaneously be zero, we have
$$
f(x)=0 \iff \sin(2x)=-\cos(2x)\iff \tan(2x)=-1=\tan\left(-\frac\pi4\right),
$$
i.e.
$$
x=\frac{2x}{2}=\frac12\left(-\frac\pi4+k\pi\right), \quad k\in \mathbb{Z}.
$$
Notice that $x\in [0,\pi]$ only if $k\in\{1,2\}$.
Hence
$$
\bbox[border: green 1px solid, 10px]{x\in \left\{\frac{3\pi}{8},\frac{7\pi}{8}\right\}.}
$$
A: $f(x) = 0 \implies \sin{2x} = - \cos{2x}$ then dividing by $\cos{2x}$ (we know $\cos{2x} \neq 0$ as $\sin{2x} \neq \cos{2x}$ and their sum is $0$) we get $\tan{2x} = -1$ and we know $\tan^{-1}(-1) = - \frac{\pi}{4}$ which allows us to find a value of x (not necessarily in the desired interval) satisfying $f(x) = 0$.
But $\tan$ is $\pi$ periodic so $\tan(2(x+ \frac{\pi}{2})) = tan(2x)$. This means whenever $x$ is a solution we can add on multiples of $\frac{\pi}{2}$ to get another solution, allowing us to find one in the desired interval.
A: Write as $\sin(2x)=-\cos(2x),$ and square both sides to get
$$\sin^2(2x)-\cos^2(2x)=0\Longrightarrow1-2\cos^2(2x)=0\Longrightarrow\cos(2x)=\pm\frac{\sqrt2}{2}.$$
Now since your restriction for $x$ is $0\le x\le\pi,$ that makes the restriction for $2x$ as $0\le 2x\le 2\pi.$
Now then $2x$ must be in the set $\{\pi/4,3\pi/4,5\pi/4,7\pi/4\}.$  See which of these work in the original equation to eliminate those which are extraneous.
A: We need $\sin2x+\cos2x=0\iff0=(\sin2x+\cos2x)^2=1+2\sin2x\cos2x=1+\sin4x$
$\implies\sin4x=-1=-\sin\dfrac\pi2=\sin\left(-\dfrac\pi2\right)$
$\implies4x=2n\pi-\dfrac\pi2$ where $n$ is any integer
