Решите уравнение: 8sin2x-2cosx-5=0

8sin2x - 5 = 0
8*(2*sinx*cosx) - 5*1 = 0
16*sinx*cosx - 5*(sin²x + cos²x) = 0
16*sinx*cosx - 5*sin²x - 5*cos²x = 0  | :cos²x≠0
16*tgx - 5*tg²x - 5 = 0
-5*tg²x + 16*tgx - 5 = 0 |*(-1)
5*tg²x - 16*tgx + 5 = 0
tgx = t
5t² - 16t + 5 = 0
D = b² - 4*a*c = 16² - 4*5*5 =  256 - 100 = 156 √D = √156 = 2√39
t1 = (16+2√39)/10 = 2(8+√39)/10 = (8+√39)/5  
t2 = (16-2√39)/10 = 2(8-√39)/10 = (8-√39)/5
tgx = (8+√39)/5
x = arccos(8+√39)/5 + pik, k ∈ Z
tgx = (8-√39)/5 
x = arccos(8-√39)/5 + pik, k ∈ Z