Question

Алгебра 10 класс
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Answer 1

$1)\; \; cos5x\cdot cos7x-sin5x\cdot sin7x=-\frac{\sqrt3}{2}\\\\cos(5x+7x)=-\frac{\sqrt3}{2}\\\\12x=\pm (\pi -\frac{\pi }{6})+2\pi n=\pm \frac{5\pi }{6}+2\pi n,\; n\in Z\\\\x=\pm \frac{5\pi }{72}+\frac{\pi n}{6}\; ,\; n\in Z\\\\2)\; \; \sqrt3\, cosx+sinx=1\\\\\frac{\sqrt3}{2}\, cosx+\frac{1}{2}\, sinx=\frac{1}{2}\\\\cos\, \frac{\pi }{6}\, cosx+sin\, \frac{\pi }{6}\, sinx=\frac{1}{2}\\\\cos(x-\frac{\pi}{6})=\frac{1}{2}\\\\x-\frac{\pi}{6}=\pm \frac{\pi}{3}+2\pi n,\; n\in Z\\\\x=\frac{\pi}{6}\pm \frac{\pi }{3+2\pi n,; n\in Z$

$3)\; \; \frac{tg\frac{\pi}{5}-tg2x}{tg\frac{\pi}{5}\cdot tg2x+1}=\sqrt3\\\\tg(\frac{\pi}{5}-2x)=\sqrt3\\\\\frac{\pi}{5}-2x=\frac{\pi}{3}+\pi n,\; n\in Z\\\\2x=-\frac{2\pi}{15}-\pi n\; ,\; n\in Z\\\\x=-\frac{\pi}{15}-\frac{\pi n}{15}\, ,\; n\in Z$