Solve for 0°<x<180° for the following equation, 8 cosec 2x cot 2x=3

8 cosec 2x . cot 2x = 3

1/sin2x . cos2x/sin2x = 3/8

cos2x/sin^{2}2x = 3/8

cos2x = 1 - 2sin^{2}2x

(1- 2sin^{2}2x)/sin^{2}2x = 3/8

1/sin^{2}2x - 2sin^{2}2x/sin^{2}2x = 3/8

1/sin^{2}2x - 2 = 3/8

1/sin^{2}2x = 3/8 + 2 = 19/8

sin^{2}2x = 8/19

sin 2x = sqrt(8/19) = 0..65

2x = sin^{-1}(0.65) = 40 degree

x = 20 degree

:)