Solving $\sin 2x + \cos 2x =-1$ 
$$\sin 2x + \cos 2x =-1$$

How would I go about solving this equation? Some hint on how to start, so I can try to figure it out on my own.
Thank you
 A: $$\sin 2x+\cos 2x=-1$$
$$2\sin x\cos x+\cos^2x-\sin^2x+\sin^2x+\cos^2x=0$$
$$2\sin x\cos x+2\cos^2x=0$$
$$\cos x(\sin x+\cos x)=0$$
$$\cos x=0\Rightarrow x=\frac{\pi}{2}+k\pi,\tan x=-1\Rightarrow x=-\frac{\pi}{4}+k\pi$$
A: Hint:
Write
$$\sin 2x=2\sin x\cos x$$
$$\cos 2x=2\cos^2x-1$$
and simplify.
A: If we consider $$A\sin 2x + B\cos 2x$$ could be of the form $$r \cos y \sin 2x+r \sin y \cos 2x=r\sin (2x+y)$$ so that $A=r\cos y, B=r\sin y$, we have $r^2=A^2+B^2$ and $\tan y=\frac BA$.
In the case in the question $r^2=2$ and $\tan y=1$ so we can have $y=\frac {\pi}4$. We reduce then to $$\sin \left(2x+\frac {\pi}4\right)=-\frac {\sqrt 2}2$$ which is then easy to solve.
A: Use sum to product for $\sin x + \sin y$ (it works also for $\cos x + \cos y$):
$$\begin{align}
\sin 2x + \cos 2x&=\sin 2x + \sin\left(\frac{\pi}{2}-2x\right)\\
&=2\sin\frac{2x+\pi/2-2x}{2}\cos\frac{2x-\pi/2+2x}{2}\\
&=2\sin\frac{\pi}{4}\cos\left(2x-\frac{\pi}{4}\right)\\
&=\sqrt 2\cos\left(2x-\frac{\pi}{4}\right)
\end{align}$$
The rest should be easy.
A: I would use the old polar-to-Cartesian $x,y$ coordinates trick, but you are already using the variable $x,$ so I'll use a $t,y$ Cartesian coordinate plane instead with the $t$ coordinate horizontal (instead of $x$) and the $y$ coordinate vertical.
Let $t = \cos (2x)$ and let $y = \sin(2x).$
Then $t^2 + y^2 = 1,$ that is, $(t,y)$ is on the unit circle centered at the origin.
We can also say that $(t,y)$ is the Cartesian coordinates of a point whose polar coordinates are $(r,\theta) = (1,2x).$
But you are given that $t + y = -1.$ Therefore $(t,y)$ is on the line
$y = -1 - t.$ Find the points where that line intersects the unit circle;
there are two such points. (A graph should make the points very easy to find in this particular case.)
Figure out the polar coordinates of those two points, and then remember
that the polar coordinates $(r,\theta)$ are also equal to $(1,2x).$
Then solve for $x.$
A: As an alternative viewpoint to momo's answer. Write,
$$\langle \cos 2x, \sin 2x \rangle \cdot \langle 1,1 \rangle=-1$$
The first vector is a unit vector, the second has length $\sqrt{1^2+1^2}$. The angle between the vectors is $2x-\frac{\pi}{4}$ or $\frac{\pi}{4}-2x$ or etc. But  it doesn't matter which one we choose because cosine is even. In fact you can check the resulting equation is exactly equal to the previous.
$$=\sqrt{1^2+1^2}\cos (2x-\frac{\pi}{4})=-1$$
A: From $\sin 2x + \cos 2x =-1$ we get with $t=2x$:
$1=\sin^2 t+\cos^2 t+2 \sin t \cos t=1+ \sin(2t)$ .
Hence $ \sin(4x)=0$.
A: It's possible to represent sum of sine and cosine of the same argument as follows
$$\sin t + \cos t = k \cdot \sin (t + \pi/4)$$
($t = 2x$ for this task), where $k$ can be found from equation above by using "sine of sum of angles" identity on the right part. So in your case initial equation will be brought to much easier one:
$$k \cdot \sin (2x + \pi/4) = -1$$
