Question

Cos2x + sin²x = 0,25 а) решите уравнение
б) найдите корень на промежутке [ -4π, -5/2π ]

Answer 1

$Cos2x+Sin^{2}x=0,25\\\\Cos^{2}x-Sin^{2}x+Sin^{2}x=\frac{1}{4}\\\\Cos^{2}x=\frac{1}{4}\\\\1)Cosx=\frac{1}{2}\\\\x_{1,2}=\pm arcCos\frac{1}{2}+2\pi n,n\in z\\\\x_{1,2}=\pm\frac{\pi }{3}+2\pi n,n\in z\\\\2)Cosx=-\frac{1}{2}$

$x_{3,4}=\pm arcCos(-\frac{1}{2})+2\pi n,n\in z\\\\x_{3,4}=\pm\frac{2\pi }{3}+2\pi n,n\in z$

$[-4\pi ;-\frac{5\pi }{2}]\\\\1)n=-1\Rightarrow x=-\frac{5\pi }{3}\\\\n=-2\Rightarrow x=-\frac{11\pi }{3} \\\\2)n=0\Rightarrow x=-\frac{\pi }{3} \\\\3)n=-1\Rightarrow x=-\frac{4\pi }{3}\\\\n=-2\Rightarrow x=-\frac{10\pi }{3}$

$4)n=0\Rightarrow x=-\frac{2\pi }{3}\\\\n=-1\Rightarrow x =-\frac{8\pi }{3}$