有a.maths 唔識做

1)
cot 2θ = 1/(tan 2θ) = (1-tan^2 θ)/(2tan θ)
1 - cot 2θ = (2tanθ - 1+tan^2 θ)/(2tan θ)
tanθ - cotθ = tanθ - 1/tanθ = (tan^2 θ - 1) / tanθ
2(tan^2 θ - 1) = (2tanθ - 1+tan^2 θ)
tan^2 θ - 2tanθ - 1 = 0
Solving, tanθ = 1- √2 or 1+√2
Hence θ = 157.5° , 337.5° , 67.5° , 247.5°

2)
Let k = sinθcosθ.
sin2θ = 2sinθcosθ = 2k
sin^4θ + cos^4θ = (sin^2 θ + cos^2 θ)^2 - 2sin^2 θcos^2 θ
= 1^2 - 2k^2 = 1 - 2k^2
Hence 1 - 2k^2 = 2k
2k^2+2k-1=0
Solving, k = (√3 - 1)/2 or (-√3-1)/2
Hence 2k = √3 - 1 or -√3-1
sin2θ = √3 - 1 or -√3-1(rejected, since sine ratio is not smaller than -1)
2θ = 47.058° , 132.94° , 492.94° , 407.06°
θ = 23.5° , 66.5° , 203° , 246°

3)
3 - 4cos2θ + cos4θ = 3 - 4cos2θ + (2cos^2 2θ - 1)
=2(cos2θ - 1)^2
Hence 2(cos2θ - 1)^2 = 1/2
(cos2θ - 1)^2 = 1/4
cos2θ - 1 = -1/2 or 1/2
cos2θ = 1/2 or 3/2(rejected, since cosine ratio is not larger than 1)
2θ = 60° , 300° , 420° , 660°
θ = 30° , 150° , 210° , 330°