đề bài: sin(2x+17pi/2) +16 = 2căn3sinxcosx +20sin^2(x/2 + pi/12)
2sin²(x/2 + π/12) = 1 - cos(x+π/6)
sin(2x+17π/2) = sin(2x+π/2 + 8π) = cos2x
có ptrình: cos2x + 16 = √3.sin2x + 10 - 10.cos(x+π/6)
<=> (1/2).cos2x - (√3/2).sin2x + 3 + 5cos(x+π/6) = 0
<=> cos(2x+π/3) + 3 + 5cos(x+π/6) = 0
<=> 2cos²(x+π/6) + 5cos(x+π/6) + 2 = 0
<=> [ cos(x+π/6) = -2 (loại)
----- [ cos(x+π/6) = -1/2
<=> [ x+π/6 = -2π/3 + k2π <=> [ x = -5π/6 + k2π
----- [ x+π/6 = 2π/3 + k2π ------ [ x = π/2 + k2π