Question

2sin^2x+5sinxcosx-8cos^2x=-1

30 баллов!!!

Answer 1

Ответ:

$2 { \sin}^{2} (x) + 5 \sin(x)\cos(x) - 8 { \cos}^{2} (x) = 1 \\ 2 { \sin }^{2} (x) + 5 \sin(x) \cos(x) - 8 { \cos }^{2} (x) = { \sin }^{2} (x) + { \cos(?) }^{2} (x) \\ { \sin }^{2} (x) + 5 \sin(x) \cos(x) - 9 { \cos}^{2} (x) = 0$

разделим на cos^2(x), не равный 0.

${tg}^{2} (x) + 5tg(x) - 9 = 0$

замена:

$tg(x) = t$

${t}^{2} + 5t - 9 = 0 \\ d = 25 + 36 = 61 \\ t1 = \frac{ - 5 + \sqrt{61}}{2} \\ t2 = \frac{ - 5 - \sqrt{61}}{2}$

$tg(x) = \frac{ - 5 + \sqrt{61}}{2} \\ x1 = arctg(\frac{ - 5 + \sqrt{61}}{2} ) + \pi \: n \\ \\ tg(x) = \frac{ - 5 - \sqrt{61}}{2} \\ x2 = - arctg(\frac{ - 5 - \sqrt{61}}{2} ) + \pi \: n$

n принадлежит Z.