Question

Сделайте пожалуйста 3

Answer 1

Ответ:

$2 { \cos }^{2} 4x + 14 { \sin}^{2} 2x - 11 = 0 \\ 2 {( \cos(4x) )}^{2} + 14 { \sin}^{2} 2x - 11 = 0 \\ 2 {(1 - 2 \sin(2x) )}^{2} + 14 { \sin }^{2} 2x - 11 = 0 \\ 2(1 - 4 \sin(2x) + 4 { \sin}^{2} 2x) + 14 { \sin }^{2} 2x - 11 = 0 \\ 2 - 8 \sin(2x) + 8 { \sin}^{2} 2x + 14 { \sin }^{2} 2x - 11 = 0 \\ 22 { \sin }^{2} 2x - 8 \sin(2x) - 9 = 0 \\ \\ \sin(2x) = t \\ \\ 22 {t}^{2} - 8 t - 9 = 0 \\ D = 64 + 792 = 856 = 214 \times 4 \\ t1 = \frac{8 + 2 \sqrt{214} }{44} = \frac{4 + \sqrt{214} }{22} \\ t2 = \frac{4 - \sqrt{214} }{22} \\ \\ \sin(2x) = \frac{4 + \sqrt{214} }{2} \\ 2x= {( - 1)}^{2} arcsin( \frac{4 + \sqrt{214} }{22} ) + \pi \: n \\ x1 = \frac{1}{2} {( - 1)}^{2} arcsin( \frac{4 + \sqrt{214} }{22} ) + \frac{\pi \: n}{2} \\ \\ \sin(2x) = \frac{4 - \sqrt{214} }{22} \\ x2 = \frac{1}{2} {( - 1)}^{2} acsin( \frac{4 - \sqrt{214} }{22} ) + \frac{\pi \: n}{2}$

n принадлежит Z.