Question

A)Решить уравнение $4cos^3X+3cosX+4\sqrt3 =4\sqrt3sin^2X$ b)найти точки на промежутке [$\frac{3\pi }{2}$; 3$\pi$]

Answer 1

$x_{1} = \frac{\pi}{2} + \pi \: n \\ x_{2} = - + \frac{5\pi}{6} + 2\pi \: n$

n€Z

€ читать "принадлежит"

б).

х=1,5π

х=2,5π

х=(2 5/6)π

Пошаговое объяснение:

$4 {cos}^{3}x + 3cosx + 4 \sqrt{3} = 4 \sqrt{3} \times {sin}^{2} x \\ 4 {cos}^{3}x + 3cosx + 4 \sqrt{3} - 4 \sqrt{3} \times {sin}^{2} x = 0 \\ 4 {cos}^{3}x + 3cosx + 4 \sqrt{3} \times (1 - {sin}^{2}x) = 0 \\$

$4 {cos}^{3}x + 3cosx + 4 \sqrt{3} \times {cos}^{2} x = 0 \\ cosx \times (4 {cos}^{2}x + 4 \sqrt{3}cosx + 3) = 0 \\ cosx = 0 \: \\ ili \\ \: 4 {cos}^{2}x + 4 \sqrt{3}cosx + 3 = 0$

cosx=0 честный случай

$x = \frac{\pi}{2} + \pi \: n$

n €Z

$4 {cos}^{2} x + 4 \sqrt{3} cosx + 3 = 0 \\ {(2cos \: x)}^{2} + 2 \times (2cosx) \times \sqrt{3} + {( \sqrt{3})}^{2} = 0$

${(2cosx + \sqrt{3})}^{2} = 0 \\ 2cosx + 3 = 0 \\ cosx = - \frac{ \sqrt{3} }{2} \\ x = - + arccos( - \frac{ \sqrt{3} }{2}) + 2\pi \: n \\ x = - + (\pi - arccos \frac{ \sqrt{3} }{2}) + 2\pi \: n \\ x = - + (\pi - \frac{\pi}{6} ) + 2\pi \: n \\ x = - + \frac{5\pi}{6} + 2\pi \: n$

n €Z

во вложении два фото