16sinx-8cos2x+7=0

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

х2 = пустому множеству