# How do you prove that cos 2x(1 + tan 2x) = 1?

Feb 2, 2015

The answer2 are: $x = k \pi$ and $\frac{\pi}{4} + k \pi$.

The equation can be written:

$\cos 2 x + \cos 2 x \tan 2 x = 1 \Rightarrow \cos 2 x + \cos 2 x \frac{\sin 2 x}{\cos 2 x} = 1 \Rightarrow \cos 2 x + \sin 2 x = 1 \Rightarrow \sin 2 x + \cos 2 x = 1$.

Now it is possible multiply both members for $\frac{\sqrt{2}}{2}$:

$\frac{\sqrt{2}}{2} \sin 2 x + \frac{\sqrt{2}}{2} \cos 2 x = \frac{\sqrt{2}}{2} \Rightarrow$

$\sin 2 x \cos \left(\frac{\pi}{4}\right) + \cos 2 x \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

$\sin \left(2 x + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

The sinus is $\frac{\sqrt{2}}{2}$ if its argument is $\frac{\pi}{4} + 2 k \pi$ or $\frac{3}{4} \pi + 2 k \pi$.

So:

$2 x + \frac{\pi}{4} = \frac{\pi}{4} + 2 k \pi \Rightarrow 2 x = 2 k \pi \Rightarrow x = k \pi$,

and

$2 x + \frac{\pi}{4} = \frac{3}{4} \pi + 2 k \pi \Rightarrow 2 x = \frac{\pi}{2} + 2 k \pi \Rightarrow x = \frac{\pi}{4} + k \pi$.