Hello, turillian!

Your set-up and differentiation is correct.

We have: .L .= .PQ .= .2.5·secθ + 2·cscθ

Then: .L' .= .2.5·secθ·tanθ - 2·cscθ·cotθ .= .0

. . . . . . . . . . . .1 . . sinθ . . . .1 . .cosθ

We have: .2.5·------·------ - 2·-----·------ .= .0

. . . . . . . . . . .cosθ .cosθ . . .sinθ .sinθ

. . Then: .2.5·sin³θ - 2·cos³θ .= .0 . . → . . 2.5·sin³θ .= .2·cos³θ

. - . . . . . .sin³θ. . . . .2

. . . And: .-------- .= .---- . . → . . tan³θ .= .0.8

. . . . . . . .cos³θ . . . 2.5

. . - . . - . . - . . - . . .____

. . Then: .tanθ .= .³√(0.8) .= .0.928317767

. . Hence: .θ .≈ .42.87°

Therefore: .L .= .2.5·sec(42.87°) + 2·csc(42.87°) .≈ .6.35 m