Trig functions (indentities) question

Question

Solve:  

 

2sin2x=cotx    , -π ≤ x ≤ 2π

Answer 1

2*sin2x = cot x


2 * 2 * sinx * cosx = cosx/sinx

4*(sinx)^2* cos x = cos x

4*(sinx)^2 * cos x - cos x = 0

cos x ( 4 * (sinx)^2 - 1 ) = 0

cos x = 0 or 4*(sinx)^2 - 1 = 0

cos x = 0 or (sin x)^2 = 1/4

cos x = 0 or sin x = +or- 1/2

For cos x = 0:

x = -90, 90, 270
= -pi/2, pi/2, (3/2)*pi


For sin x = +or- 1/2:

x= -150, -30, 30, 150, 210, 330

= (-5/6)*pi, -pi/6, pi/6, (5/6)*pi, (7/6)*pi, (11/6)*pi