Expanding the left side as the fifth power of a binomial, equating the imaginary parts on both sides of the equation, and then replacing sin272∘ with 1−cos272∘
isin72∘(5cos472∘+10cos272∘i2sin272∘+i4sin472∘)=0i
16cos472∘−12cos272∘+1=0
Solving this quadratic in cos272∘, we get
cos272∘=12±√8032
so
cos272∘=6±2√516=(√5±1)242
cos72∘=±√5±14
where we can choose the correct value of the four possible values by noting that, because 72° is between 45° and 90°, cos72∘ must lie between 1/√2 and 0. Because cos72∘ is positive, we choose the "+" before the fraction, and because cos72∘ is less than 1/√2, which in turn is less than √5+14, we choose the "-" in the numerator:
cos72∘=√5−14Constructing a regular pentagon
So we can construct cos72∘. For example, the diagonal of a 1-by-2 rectangle is √5. We could cut off one unit from a segment of length √5, then divide the segment of length √5−1 into four pieces of length √5−14. (Or we could construct the appropriate solution to the equation 4x2+2x−1=0. See my post on Solving quadratic equations via geometric construction.)Construct a unit circle centered at O, and construct a radius ¯OA. Construct the point B on ¯OA so that ¯OB has length cos72∘. If C is a point on the circle so that ¯BC is perpendicular to ¯OA, then ∠COA is a 72° angle, and both A and C are vertices of a regular pentagon inscribed in the circle.