how to solve (1-sin2x) / (1+sin2x) = ctg [pi/4 + x]?

Question

We are writing without brackets..... ctg =cot sin2x=sin(2x)

It should be cot^2[pi/4 + x] not cot[pi/4 +x]

Answer 1

If all fails, converting everything to sines and cosines should always work.

(1-sin2x)/(1+sin2x) = cot[pi/4 + x]
[1-2sin(x)*cos(x)]/[1+2sin(x)*cos(x)] = 1/tan(pi/4 + x)
[1-2sin(x)*cos(x)]/[1+2sin(x)*cos(x)] = [1-tan(x)]/[tan(x)+1]
[1-2sin(x)*cos(x)]/[1+2sin(x)*cos(x)] = [cos(x)-sin(x)]/[cos(x)+sin(x)]
[1-2sin(x)*cos(x)]*[cos(x)+sin(x)] = [cos(x)-sin(x)]*[1+2sin(x)*cos(x)]

cos(x)+sin(x)-2sin(x)*[cos(x)]^2-2[sin(x)]^2*cos(x) =
cos(x)-sin(x)+2sin(x)*[cos(x)]^2-2[sin(x)]^2*cos(x)

sin(x)-2sin(x)*[cos(x)]^2 = -sin(x)+2sin(x)*[cos(x)]^2
2sin(x)-4sin(x)*[cos(x)]^2 = 0
sin(x)*(1-2[cos(x)]^2) = 0

sin(x) = 0, x = nπ
cos(x) = +-sqrt(2)/2, x = π/4 + nπ/2
However, cot(x+π/4) does not exist when n is odd for the second case.

Hence, the solution is
x = {nπ, π/4 + nπ}

Answer 2

Does ctg mean cot?
Does sin2x mean sin(2x)?

You need to state your problems clearly and unambiguously.