A curve passes through the point (1, π/6). Let the slope of the curve at each point (x, y) be yx + sec(yx), x > 0. Then the equation of the curve is:

Given that the slope of the curve at each point (x, y) is yx + sec(yx), we have:
dy/dx = y/x + sec(y/x)
Let v = y/x. Then y = vx, and dy/dx = v + x(dv/dx).
Substituting this into the given equation, we get:
v + x(dv/dx) = v + sec(v)
x(dv/dx) = sec(v)
dv/sec(v) = dx/x
cos(v) dv = dx/x
Integrating both sides, we get:
∫cos(v) dv = ∫dx/x
sin(v) = ln|x| + C
Since x > 0, we can write:
sin(v) = ln(x) + C
Substituting v = y/x, we get:
sin(y/x) = ln(x) + C
The curve passes through the point (1, π/6). Substituting x = 1 and y = π/6, we get:
sin(π/6) = ln(1) + C
1/2 = 0 + C
C = 1/2
Therefore, the equation of the curve is:
sin(y/x) = ln(x) + 1/2