<s>[INST] How do I solve this equation? $\displaystyle \large \cos 2x + 1 - \sin 2x=\frac{2 \cos 2x \cos x}{\cos x + \sin x}$
I've been trying for a long time but I can't get it.
 [/INST] Using the identities 
$$
 {\cos 2x=\cos^2 x -\sin^2  x},\quad  \cos 2x=2\cos^2 x-1,\quad \sin 2x=2\sin x\cos x
$$
We have, if $\cos x+\sin x\ne0$
$$\eqalign{
{2 {\cos 2x} \cos x\over \cos x+\sin x}
&={2(  {\cos^2 x-\sin^2 x}) \cos x\over \cos x+\sin x}\cr
&={2(\cos x+\sin x)(\cos x-\sin x) \cos x\over \cos x+\sin x}\cr
&={2 (\cos x-\sin x) \cos x }\cr
&=2\cos^2 x -2\sin x\cos x\cr
&=\cos 2x +1 -\sin 2x.
}
$$
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