Question

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

Answer 1

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