8sin^2x+6sin(pi/2+x)=9
Надеюсь все видно....
Sin(π/2 + x) = cosx 8 (1-cos²x) + 6cosx = 9 8- 8cos²x + 6cosx -9=0 -8cos²x + 6cosx - 1=0 8cos²x - 6cosx + 1=0 cosx = t; ItI≤1 8t² - 6t + 1=0 D₁ = 9-8=1 t₁ = (3+1)/8 = 1/2 t₂ = (3-1)/8 = 1/4 cosx = 1/2 cosx = 1/4 x = π/3 + 2πn x = arccos1/4 + 2πn