Question

Найдите решение уравнения
4sinx=5-4cos^2x

Answer 1

$4\sin{x}=5-4\cos^2{x}\\4\sin{x}=5-4(1-\sin^2{x})\\4\sin{x}-5+4-4\sin^2{x}=0\\4\sin^2{x}-4\sin{x}+1=0\\t=\sin{x}\\\\4t^2-4t+1=0\\\displaystyle t=\frac{1}{2}\\\sin{x}=\frac{1}{2}\\x=\displaystyle\begin{cases}\frac{\pi}{6} + 2\pi n\\\frac{5\pi}{6}+2\pi n, n \in \mathbb Z\end{cases}$