Question

Решите уравнение 5sin2x-4sinxcosx+3cos2x=0

Answer 1

5sin2x - 4sinxcosx + 3cos2x=0
5sin2x - 2sin2x + 3cos2x = 0
3sin2x + 3cos2x = 0  делим на 3
cos^2(x) + 2sinxcosx - sin^2(x) = 0  делим на cos^2(x)
1 + 2tgx - tg^2(x) = 0
tg^2(x) - 2tgx - 1 = 0  назначим tg(x) = t
t^2 - 2t - 1 = 0  решим
D = 4 + 4 = 8
t1 = (2 + 2*$sqrt{2}$) / 2 = 1 + $sqrt{2}$
t2 = (2 - 2*$sqrt{2}$) / 2 = 1 - $sqrt{2}$

tgx = 1 + $sqrt{2}$
x = arctg(1 + $sqrt{2}$) + $pi$*n  где n ∈ Z

tgx = 1 - $sqrt{2}$
x = arctg(1 - $sqrt{2}$) + $pi$*n  где n ∈ Z