Question

Математика помогите пжпж!

Answer 1

Ответ:

${sin}^{2} 2x + {cos}^{2} x = 1 \\ 4 {sin}^{2} x \: {cos}^{2} x + 1 - {sin}^{2} x = 1 \\ {sin}^{2} x (4 {cos}^{2} x - 1) = 0$

${sin}^{2} x = 0 \\ x = \pi \: n,n∈Z$

$4 {cos}^{2} x - 1 = 0 \\ cosx = \frac{1}{2}$

$x=±arccos \frac{1}{2} + 2\pi \: n ,n∈Z \\ x=± \frac{\pi}{3} + 2\pi \: n ,n∈Z$

Ответ :

$x = \pi \: n,n∈Z \\ x=± \frac{\pi}{3} + 2\pi \: n ,n∈Z$