Trigonometry problem

$\begin{aligned} f(x) & = \cos 4x - 4 \sin 2x + 2 (\cos x + \sin x)^4 \\ & = 1 - 2 \sin^2 2x - 4 \sin 2x + (\cos^2 x + 2\sin x \cos x + \sin^2 x)^2 \\ & = 3 - 2(\sin^2 2x + 2 \sin 2x + 1) + 2(\sin 2x + 1)^2 \\ & = 3 - 2(\sin 2x + 1)^2 + 2(\sin 2x + 1)^2 \\ & = \boxed 3 \end{aligned}$

Let's expand $(\cos x + \sin x)^2 = \cos^2 x + \sin^2 x + 2 \cos x \sin x = 1 + \sin 2x$ . (Using double angle formula for $\sin 2x$ )

If we square this again, we get $(1 + \sin 2x)^2 = 1 + 2 \sin 2x + \sin^2 2x$

Thus $2 (\cos x + \sin x)^4 = 2 + 4 \sin 2x + 2 \sin^2 2x$ .

Also, $\cos 4x = \cos^2 2x - \sin^2 2x$ . (Using double angle formula for $\cos 4x$ )

Now consider the sum as a whole:

$\cos 4x - 4 \sin 2x + 2 (\cos x + \sin x)^4 = \cos^2 2x - \sin^2 2x - 4 \sin 2x + 2 + 4 \sin 2x + 2 \sin^2 2x = \ldots$

(Combining and rearranging terms)

$= (\cos^2 2x + \sin^2 2x) + (- 4 \sin 2x + 4 \sin 2x) + 2$

$= 1 + 0 + 2 = 3$

As the answer is a constant, you can try substitute x=0, and get 3.