Question

21sin^2 x+sin2x−15cos^2 x=2

Answer 1

Ответ:

$21 { \sin }^{2} x + \sin(2x) - 15 { \cos }^{2} x = 2 \\ 21 { \sin }^{2} x + 2 \sin(x) \cos(x) - 15 { \cos }^{2} x = 2 { \sin}^{2} x + 2 { \cos }^{2} x \\ 19 { \sin}^{2} x + 2\sin(x) \sin(x) - 17 { \cos}^{2} x = 0$

разделим на cos^2x, не равный 0.

$19 {tg}^{2} (x) + 2tg(x) - 17 = 0 \\ \\ tg(x) = t \\ 19 {t}^{2} + 2t - 17 = 0 \\ d = 4 + 1292 = 1296 \\ t1 = \frac{ - 2 + 36}{19 \times 2} = \frac{34}{2 \times 19} = \frac{17}{19} \\ t2 = \frac{ - 38}{2 \times 19} = - 1 \\ \\ tg(x) = \frac{17}{19} \\ x1 = arctg( \frac{17}{19} ) + \pi \: n \\ \\ tg(x) = - 1 \\ x2 = - \frac{\pi}{4} + \pi \: n$

n принадлежит Z.