Question

$9^{2sin^2x+cosx-1} =1$

решите плиз

Answer 1

$9^{2sin^2x + cosx - 1 } = 1\\ 9^{2sin^2x + cosx - 1} = 9^0 \\ 2sin^2x + cosx -1 = 0 \\ 2(1-cos^2x) + cosx - 1 = 0 \\ 2 - 2cos^2x + cosx - 1 = 0 \\ -2cos^2x + cosx + 1 = 0 \\ 2cos^2x - cosx - 1 = 0 \\ cosx = t \\ 2t^2 - t - 1 = 0 \\ (2-1-1= 0) =\ \textgreater \ t1 = 1, t2 = c/a = -1/2 \\ cosx = 1 \\ x = 2pik \\ cosx = -1/2 \\ x1 = +-2pi/3 + 2pin$