Question

Знатоки алгебры, помогите, пожалуйста. Решите уравнения.
1)
$sin {}^{4} \frac{x}{2} - cos {}^{4} \frac{x}{2} = \frac{1}{4}$
2)
$cos {}^{2} x - cos {}^{2}2 x + \cos {}^{2} (3x) - \cos {}^{2} (4x) = 0$
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Answer 1

Ответ:

$1)\ \ sin^4\dfrac{x}{2}-cos^4\dfrac{x}{2}=\dfrac{1}{4}\\\\\Big(\underbrace{sin^2\dfrac{x}{2}-cos^2\dfrac{x}{2}}_{-cos(2\cdot \frac{x}{2})}\Big)\Big(\underbrace {sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}}_{1}\Big)=\dfrac{1}{4}\\\\\\-cosx=\dfrac{1}{4}\ \ ,\ \ \ cosx=-\dfrac{1}{4}\ \ ,\\\\\\x=\pm arccos\Big(-\dfrac{1}{4}\Big)+2\pi n\ ,\ n\in Z\\\\x=\pm \Big(\pi -arccos\dfrac{1}{4}\Big)+2\pi n\ ,\ n\in Z$

$2)\ \ cos^2x-cos^22x+cos^23x-cos^24x=0\\\\\\\dfrac{1+cos2x}{2}-\dfrac{1+cos4x}{2}+\dfrac{1+cos6x}{2}-\dfrac{1+cos8x}{2}=0\ \Big|\cdot 2\\\\\\1+cos2x-1-cos4x+1+cos6x-1-cos8x=0\\\\(cos2x-cos4x)+(cos6x-cos8x)=0\\\\2sin3x\cdot sinx+2sin7x\cdot sinx=0\\\\2sinx\cdot (sin3x+sin7x)=0\\\\sinx\cdot 2sin5x\cdot cos2x=0\\\\a)\ \ sinx=0\ \ ,\ \ x=\pi n\ ,\ n\in z\\\\b)\ \ sin5x=0\ \ ,\ \ 5x=\pi k\ ,\ \ x=\dfrac{\pi k}{5}\ ,\ k\in Z\\\\\{\pi n\}\in \{\frac{\pi k}{5}\}\ \ pri\ \ k=5p\ ,\ p\in Z$

$c)\ \ cos2x=0\ \ ,\ \ 2x=\dfrac{\pi}{2}+\pi m\ \ ,\ \ x=\dfrac{\pi}{4}+\dfrac{\pi m}{2}\ ,\ m\in Z\\\\\\Otvet:\ \ x=\dfrac{\pi k}{5}\ ,\ x=\dfrac{\pi}{4}+\dfrac{\pi m}{2}\ ,\ \ k,m\in Z$