Question

(2 sin^2 4x - 3cos 4x) *√tgx=0

Answer 1

$(2sin^24x - 3cos4x) cdot sqrt{tgx} = 0$
ОДЗ:
$tgx geq 0 \ pi n leq x leq dfrac{ pi }{2}+ pi n, n in Z$
$tgx = 0 2sin^24x - 3cos4x = 0 \\ boxed{ x = pi n, n in Z } \ \ 2 - 2cos^24x - 3cos4x = 0 \ \ 2cos^24x + 3cos4x - 2 = 0$
Пусть $t = cos4x, t in [-1; 1]$
[$2t^2 + 3t - 2 = 0 \ \ D = 9 + 4 cdot 2 cdot 2 = 25 = 5^2$
$t_1 = dfrac{-3 + 5 }{4} = dfrac{1}{2}$
$t_2 = dfrac{-3 - 5}{2} = -2$ - посторонний корень
Обратная замена:
$cos4x = dfrac{1}{2} \ \ 4x = pm dfrac{ pi }{3} + 2 pi n, n in Z \ \ x = pm dfrac{ pi }{12} + dfrac{ pi n}{2}, n in Z$

$pi n leq pm dfrac{ pi }{12} + dfrac{ pi n}{2} leq pi n + dfrac{ pi }{2}, n in Z \ \ 12n leq pm 1 + 6n leq 12n + 6, n in Z \ \ 0 leq pm 1 - 6n leq 6, n in Z \ \ n = -1; 0.$

$x = pm dfrac{ pi }{12} \ \ x = - dfrac{ pi }{12} + dfrac{ pi }{2} = - dfrac{ pi }{12} +dfrac{6 pi }{12} = dfrac{5 pi }{12}$