Question

Помогите пожалуйста 2sin²x+cosx-1=0, K ∈ Z

Answer 1

Ответ:

$-\frac{\pi}{3}+2\pi k,~~\frac{\pi}{3}+2\pi k,~~2\pi n,~~k,n\in Z$

Объяснение:

$2\sin^2{x}+\cos{x}-1=0 \\ \\ 2(1-\cos^2{x})+\cos{x}-1=0 \\ \\ 2-2\cos^2{x}+\cos{x}-1=0 \\ \\-2\cos^2{x}+\cos{x}+1=0 ~~~~~|\cdot(-1) \\ \\2\cos^2{x}-\cos{x}-1=0 \\\\ \cos{x}=t,~~~-1\leq t\leq 1 \\ \\ 2t^2-t-1=0 \\ \\ D=(-1)^2-4\cdot 2\cdot (-1)=1+8=9>0 \\ \\ t_{1,2}=\frac{-(-1)\pm \sqrt{9}}{2\cdot 2} =\frac{1\pm 3}{4}$

$\left[\begin{array}{c}{t_1=-\frac{1}{2} }&{t_2=1}\end{array}$

$\left[\begin{array}{c}{\cos{x}=-\frac{1}{2} }&{\cos{x}=1}\end{array}$

$\left[\begin{array}{c}{x=\pm \arccos(-\frac{1}{2})+2\pi k,~k\in Z }&{x=2\pi n, n \in Z~~~~~~~~~~~~~~~~~~~~~}\end{array}$

$\left[\begin{array}{c}{x=\pm(\pi- \arccos\frac{1}{2})+2\pi k,~k\in Z }&{x=2\pi n, n \in Z~~~~~~~~~~~~~~~~~~~~~}\end{array}$

$\left[\begin{array}{c}{x=\pm(\pi- \frac{\pi}{3})+2\pi k,~k\in Z }&{x=2\pi n, n \in Z~~~~~~~~~~~~~~~~}\end{array}$

$\left[\begin{array}{c}{x=\pm\frac{2\pi}{3}+2\pi k,~k\in Z }&{x=2\pi n, n \in Z~~~~~~~~~~~}\end{array}$