simple trigonometric functions how would you solve this 
$$\cos^2 x - 2\sin x \cos x - \sin^2 x = 0$$
I tried to simplify it but I got $\cos 2x - \sin 2x$. I can't simplify that further.
 A: Based on what you have done,
\begin{align*}
\cos^2x - 2\sin x\cos x - \sin^2x & = 0\\
\cos^2x - \sin^2x - 2\sin x\cos x & = 0\\
\cos(2x) - \sin(2x) & = 0\\
\cos(2x) & = \sin(2x)\\
1 & = \tan(2x)
\end{align*}
Observe that if you want values of $x$ in the interval $[0, 2\pi)$ that satisfy the equation, then $0 \leq 2x < 4\pi$.  Thus, we must find values of $2x$ in the interval $[0, 4\pi)$ such that $\tan(2x) = 1$.  They are 
\begin{align*}
2x & = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}\\
x & = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}
\end{align*}
Since $\tan(2x)$ has period $\dfrac{\pi}{2}$, the general solution is 
$$x = \frac{\pi}{8} + n\frac{\pi}{2}, n \in \mathbb{Z}$$
A: From what you have reached.
$$\begin{align}(\cos 2x - \sin 2x)^2 = 0  &\Rightarrow 2\sin2x\cos2x = 1 \Rightarrow \sin 4x = 1 \\ &\Rightarrow x = \frac{\pi}{8} + \frac{\pi k}{2},\  k \in \mathbb{Z}\end{align}$$
A: You have that $\cos2x-\sin2x=0$, or in other words, $\cos2x = \sin2x$. Where on the unit circle are the cosine and sine the same? That's the value of $2x$.
A: $\begin{align}(\cos 2x-\sin 2x)=0 & \implies\dfrac{1}{\sqrt{2}}\cos 2x-\dfrac{1}{\sqrt{2}}\sin 2x=0\\&\implies \sin\left(\dfrac{\pi}{4}-2x\right)=0\end{align}$.
