Question

Решите уравнения, найдите корни, принадлежащие промежутку: $(2cos^{2} x+sinx-2)\sqrt{5tgx} =0$ $[\pi ; \frac{5\pi }{2} ]$

Answer 1

$(2cos^{2} x+sinx-2)\sqrt{5tgx} =0 \\ [\pi ; \frac{5\pi }{2} ]$
$(2(1 - { \sin }^{2} x) + \sin(x) - 2) = 0 \\ \sqrt{5} \sqrt{ \tg(x) } = 0$
$- 2 \ {sin}^{2} (x) + \sin( x ) = 0 \\ \sin(x)( - 2sin(x) + 1)= 0 \\ x = \pi \times k \\ x = \frac{\pi}{6} + \pi \times k = \frac{(6k + 1)\pi}{6}$
Ответ:
$x = \pi \\ x = 2\pi \\ x = \frac{7\pi}{6} \\ x = \frac{13\pi}{6}$