Find an equation in rectangular coordinates for the equation given in spherical coordinates: $\phi=\pi/6$
Equation must be such that $z \ge 0$.
Here is what I did:
$z = \rho\cos\phi$
$z = \cos{\pi/6}\sqrt{x^2 +y^2 + z^2}$
and since z must be greater than or equal to zero:
z = $| \cos{\pi/6}\sqrt{x^2 +y^2 + z^2} |$
On
What you have is an equation in $x,$ $y,$ and $z$ whose solution is the desired surface, so it's technically correct, provided that we interpret $\cos \pi/6$ as $\cos(\pi/6)$ and not $(\cos\pi)/6.$ You might also want to format your expression carefully so that someone doesn't think you meant $\cos\left(\frac{\pi\sqrt{x^2+y^2+z^2}}{6}\right)$ or even $\cos\left(\frac{\pi}{6\sqrt{x^2+y^2+z^2}}\right).$
I think this works well to remove those ambiguities: $z = \left\lvert\sqrt{x^2+y^2+z^2}\cos\frac{\pi}{6}\right\rvert.$
You could still do a lot to simplify the expression, however. As has already been pointed, out, the absolute value is redundant since $\cos\frac{\pi}{6} > 0$ and since $\sqrt{x^2+y^2+z^2}$ is non-negative by definition of the square root function. But the fact that you have a square root suggests you might try to see what happens if you square both sides. If you do that, and if you use the well-known value of $\cos\frac{\pi}{6},$ you might find that things simplify quite a bit (although the danger of squaring is that it sometimes allows solutions you don't want, and in this case it forces you to explicitly require that $z \geq 0$).
Please note how your expression of $z$ is already positive, you don't need absolute values. Else, your expression is correct.