So the question goes like this
Solve the following trigonometric equation finding all solutions for X in the inerval [0, 2pi]
2cos(2x)csc^2(x)=2cos(2x)
Thanks in advance.
Another way of looking at it: obviously if cos(x)= 0 then both sides are 0 so cos(x)= 0 gives solutions. If cos(x) is NOT 0, we can divide by 2cos(x) to get so that csc(x)= 1 and csc(x)= -1 gives solutions.