Пошаговое объяснение:
cos3x≤1/2
π/3+2πn≤3x≤5π/3+2πn ,n∈z
π/9+2πn/3≤x≤5π/9+2πn/3,n∈z
Решение.
Тригонометрическое неравенство :
[tex]\bf cos\, 3x\leq \dfrac{1}{2}\ \ \ \Rightarrow \ \ \ \dfrac{\pi }{3}+2\pi k\leq 3x\leq 2\pi -\dfrac{\pi }{3}+2\pi k\ \ ,\ k\in Z\\\\\\\dfrac{\pi }{3}+2\pi k\leq 3x\leq \dfrac{5\pi }{3}+2\pi k\ \ ,\ k\in Z\\\\\\\dfrac{\pi }{9}+\dfrac{2\pi k}{3}\leq x\leq \dfrac{5\pi }{9}+\dfrac{2\pi k}{3}\ \ ,\ k\in Z\\\\\\x\in \Big[\ \dfrac{\pi }{9}+\dfrac{2\pi k}{3}\ ;\ \dfrac{5\pi }{9}+\dfrac{2\pi k}{3}\ \Big]\ \ ,\ k\in Z[/tex]
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Answers & Comments
Пошаговое объяснение:
cos3x≤1/2
π/3+2πn≤3x≤5π/3+2πn ,n∈z
π/9+2πn/3≤x≤5π/9+2πn/3,n∈z
Решение.
Тригонометрическое неравенство :
[tex]\bf cos\, 3x\leq \dfrac{1}{2}\ \ \ \Rightarrow \ \ \ \dfrac{\pi }{3}+2\pi k\leq 3x\leq 2\pi -\dfrac{\pi }{3}+2\pi k\ \ ,\ k\in Z\\\\\\\dfrac{\pi }{3}+2\pi k\leq 3x\leq \dfrac{5\pi }{3}+2\pi k\ \ ,\ k\in Z\\\\\\\dfrac{\pi }{9}+\dfrac{2\pi k}{3}\leq x\leq \dfrac{5\pi }{9}+\dfrac{2\pi k}{3}\ \ ,\ k\in Z\\\\\\x\in \Big[\ \dfrac{\pi }{9}+\dfrac{2\pi k}{3}\ ;\ \dfrac{5\pi }{9}+\dfrac{2\pi k}{3}\ \Big]\ \ ,\ k\in Z[/tex]