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