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:
$cos4x = \dfrac{1}{2} \\ \\ 4x = \pm \dfrac{ \pi }{3} + 2 \pi n, \ n \in Z \\ \\ x = \pm \dfrac{ \pi }{12} + \dfrac{ \pi n}{2}, \ n \in Z$

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

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