Question

16sinx-8cos2x+7=0

Answer 1

$<var>16sinx-8cos2x+7=0\ 16sinx-8(1-2sin^2x)+7=0\ 16sinx-8+16sin^2x+7=0\ 16sin^2x+16sinx-1=0\ sinx = y\ 16y^2+16y-1 = 0\ D = 256 -4*16*(-1) = 256+64 = 320\ y1 =frac{-16+8sqrt{5}}{32} = frac{-8(2-sqrt{5})}{32} = - frac{2-sqrt{5}}{4} \ y2= - frac{2+sqrt{5}}{4} \ sinx_1 = - frac{2-sqrt{5}}{4} \ x_1 = (-1)^narcsin(- frac{2-sqrt{5}}{4})+pi n, neZ\ \ sinx_2 = - frac{2+sqrt{5}}{4}$ 

х2 = пустому множеству