Question

Помогите решить пример на фото пожалуйста
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Answer 1

Ответ:

$-\dfrac{\pi}{18}+\dfrac{n\pi}{6},\;n\in Z$

Объяснение:

$\sqrt{3}cos^23x+sin6x-\sqrt{3}sin^23x=0\\\sqrt{3}(cos^23x-sin^23x)+sin6x=0\\\sqrt{3}cos6x+sin6x=0\\2\left(sin\dfrac{\pi}{3}cos6x+sin6xcos\dfrac{\pi}{3}\right)=0\\sin\left(6x+\dfrac{\pi}{3}\right)=0\\6x+\dfrac{\pi}{3}=n\pi,\;n\in Z\\x=-\dfrac{\pi}{18}+\dfrac{n\pi}{6},\;n\in Z$

Уравнение решено!

Answer 2

$\sqrt3cos^23x+sin6x-\sqrt3sin^23x=0\\\\\sqrt3cos^23x+2\, sin3x\cdot cos3x-\sqrt3sin^23x=0\ |:cos^23x\ne 0\\\\\sqrt3+2tg3x-\sqrt3tg^23x=0\\\\\sqrt3tg^23x-2tg3x-\sqrt3=0\\\\t=tg3x\ \ ,\ \ \sqrt3t^2-2t-\sqrt3=0\ \ ,\ \ D/4=1+3=4\ ,\ \ t_{1,2}=\dfrac{1\pm 2}{\sqrt3}\\\\a)\ \ tg3x=-\dfrac{1}{\sqrt3}\ \ ,\ \ 3x=-\dfrac{\pi}{6}+\pi n\ \ ,\ \ x=-\dfrac{\pi}{18}+\dfrac{\pi n}{3}\ ,\ n\in Z\\\\b)\ \ tg3x=\dfrac{3}{\sqrt3}=\sqrt3\ ,\ \ 3x=\dfrac{\pi}{3}+\pi k\ ,\ \ x=\dfrac{\pi}{9}+\dfrac{\pi k}{3}\ ,\ k\in Z$

$\left[\begin{array}{l}x=-\dfrac{\pi}{18}+\dfrac{\pi n}{3}\ ,\ n,\in Z\\x=\dfrac{\pi}{9}+\dfrac{\pi k}{3}\ ,\ k\in Z\ .\end{array}\right\ \ \ \Rightarrow \ \ \ x=-\dfrac{\pi}{18}+\dfrac{\pi m}{6}\ ,\ m\in Z\\\\\\Otvet:\ \ x=-\dfrac{\pi }{18}+\dfrac{\pi m}{6}\ ,\ m\in Z\ .$