Wasn't the exact same question
View attachment 29803
Only difference really was part iii but I'm sure what had to be done there could have also been applied to iii in my exam
Here's the link
finding exact value is unnecessary. but u could do it.
4x^2+2x-1 = 0 is equivalent to proving
x^2 + 1/2x - 1/4 = (x- cos2pi/5)(x-cos4pi/5), for x = cos2pi/5, x = cos 4pi/5.
2(cos2pi/5 + cos4pi/5) + 1 = 0 (original eq), cos2pi/5 + cos4pi/5 = -1/2
(cos2pi/5 cos4pi/5 )^2 x 1 = 1.
cos2pi/5 cos4pi/5 = -1 or 1.
cos2pi/5 1st quadrant 4pi/5 2nd quadrant.
therefore -1.
hence lhs = rhs.
(messed up somewhere oops.)