3csc^2x-4=0 solve [0,2pi}

I'm assuming that you meant "3 (csc x) ^2 - 4 = 0." Your "csc^2x" is ambiguous.

If 3 (csc x) ^2 - 4 = 0, then 3[1/sin x]^2 - 1 = 0 is equivalent. We need to solve this for x.

First, rewrite this as 3[1/sin x]^2 = 4.

Next, divide both sides by 3: [1/sin x]^2 = 4/3

This quadratic equation has two roots, one positive and one negative.

Supposing [1/sin x]^2 = 4/3, then 1/sin x = 2/sqrt (3). Inverting both sides, sin x = sqrt (3) / 2. Find x by finding the inverse function:

x=arcsin sqrt (3) / 2 so x = 60 degrees, or pi/3 radians.

Now find the other root or roots.