Решите уравнение: 
8sin2x-2cosx-5=0 

8sin2x - 5 = 08*(2*sinx*cosx) - 5*1 = 016*sinx*cosx - 5*(sin²x + cos²x) = 016*sinx*cosx - 5*sin²x - 5*cos²x = 0  | :cos²x≠016*tgx - 5*tg²x - 5 = 0-5*tg²x + 16*tgx - 5 = 0 |*(-1)5*tg²x - 16*tgx + 5 = 0tgx = t5t² - 16t + 5 = 0D = b² - 4*a*c = 16² - 4*5*5 =  256 - 100 = 156 √D = √156 = 2√39t1 = (16+2√39)/10 = 2(8+√39)/10 = (8+√39)/5  t2 = (16-2√39)/10 = 2(8-√39)/10 = (8-√39)/5tgx = (8+√39)/5x = arccos(8+√39)/5 + pik, k ∈ Ztgx = (8-√39)/5 x = arccos(8-√39)/5 + pik, k ∈ Z