Question

при каких значениях х принадлежит [0;П] 2cos2x равняется -1​

Answer 1

Ответ:

x=5π/6

Объяснение:

2cos2x = -1

cos2x = -1/2     [0;π]

2x = ±π/3+2πk;   k∈Ζ

x = ±π/6+πk; [0;π]  k=1

x= -π/6+π=5π/6

Answer 2

$2Cos2x=-1\\\\Cos2x=-\dfrac{1}{2} \\\\2x=\pm arcCos(-\frac{1}{2})+2\pi n,n\in Z\\\\2x=\pm \dfrac{2\pi }{3}+2\pi n,n\in Z\\\\x=\pm \frac{\pi }{3}+\pi n,n\in Z\\\\\left[\begin{array}{ccc}x_{1}=-\dfrac{\pi }{3}+\pi n,n\in Z \\x_{2} =\dfrac{\pi }{3}+\pi n,n\in Z \end{array}\right$

$1)x=-\dfrac{\pi }{3}+\pi n,n\in Z\\\\0\leq -\dfrac{\pi }{3}+\pi n\leq \pi |\cdot \dfrac{3}{\pi }\\\\0\leq -1+3n\leq 3\\\\1\leq 3n\leq 4\\\\\dfrac{1}{3}\leq n\leq 1\frac{1}{3} \\\\n=1 \ \Rightarrow \ x=-\dfrac{\pi }{3} +\pi=\dfrac{2\pi }{3}\\\\\boxed{x=\dfrac{2\pi }{3}} \\\\\\2)x=\dfrac{\pi }{3}+\pi n,n\in Z\\\\0\leq \dfrac{\pi }{3}+\pi n\leq \pi |\cdot \dfrac{3}{\pi }\\\\0\leq 1+3n\leq 3\\\\-1\leq 3n\leq 2\\\\-\dfrac{1}{3}\leq n\leq \dfrac{2}{3}$

$n=0 \ \Rightarrow \ \boxed{x=\dfrac{\pi }{3}}$