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Problem 725

AIME late

Algebra Difficulty 5.3 Find the answer

5.4 $2(\cos 4 x-\sin x \cos 3 x)=\sin 4 x+\sin 2 x$ .

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

5.4 $2(\cos 4 x-\sin x \cos 3 x)=\sin 4 x+\sin 2 x$ .

Your answer

Official solution

5.4 Let's rewrite the equation as

$\cos 4 x=\sin x \cos 3 x+0.5(\sin 4 x+\sin 2 x)$ .

Transform the right side of the equation:

$\sin x \cos 3 x+0.5 \cdot 2 \sin 3 x \cos x=\sin 4 x$

Thus, $\cos 4 x=\sin 4 x$ , from which $\operatorname{tg} 4 x=1 ; 4 x=\frac{\pi}{4}+\pi k, x=$ $=\frac{\pi}{16}(4 k+1)$ (division by $\cos 4 x$ is possible since the values of $x$ for which $\cos 4 x=0$ are not solutions to the equation).

Answer: $\quad x=\frac{\pi}{16}(4 k+1), k \in Z$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.