tan2x = 8cos^2x - cotx

Question

0 < x < pi/2. The answers are pi/24 and 5pi/24 but I don't understand how to get to the answer. It involves trig formulas but I don't know which ones to use or how to use them.

Answer 1

tan2x = 8cos^2 x - cotx

=> 2tanx/(1-tan^2 x) = 8cos^2 x - cotx

=> 2tanx = (1-tan^2 x) (8cos^2 x - cotx)

=> 2tanx = 8cos^2 x - cotx - tan^2 x * 8cos^2 x + tan^2 x * cotx

=> 2tanx = 8cos^2 x - cotx - 8 sin^2 x + tanx {since tan=sin/cos and tan = 1/cot}

=> 2tanx = 8cos2x  - cotx + tanx {since, cos^2 x - sin^2x = cos2x}

=> 2tanx - tanx + cotx = 8 cos2x

=> tanx + cotx = 8 cos2x

=> sinx/cosx + cosx/sinx = 8 cos2x

=> (sin^2 x + cos^2x)/(cosx * sinx) = 8 cos2x

=> 1 = cosx*sinx * 8 cos2x

=> 1 = 4sin2x * cos2x

=> 1 = 2 sin4x {since 2sinxcosx = sin2x}

=> sin4x = 1/2

So, 4x = arcsin(1/2)

or 4x = pi/6 or 4x = 5pi/6

or x = pi/24 or x = 5pi/24