Question

Помогите пожалуйста!
А17

Answer 1

$cos\dfrac{2 \pi x}{3}sin\dfrac{\pi x}{6}=1+sin\dfrac{2 \pi x}{3} cos \dfrac{\pi x}{6} \\ cos\dfrac{2 \pi x}{3}sin\dfrac{\pi x}{6}-sin\dfrac{2 \pi x}{3} cos \dfrac{\pi x}{6}=1\\ sin\dfrac{\pi x}{2}=-1\\ \dfrac{\pi x}{2}=-\dfrac{\pi }{2}+2 \pi k\\ \boxed{x=-1+4k; \ k \in Z}$

Answer 2

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