TMUA 2018 Paper 2 Question 4

First pin down c. On 0 < x < 32π the cosine falls from 1 to −1 over (0, π) and rises from −1 back to 0 over (π,  32π), so cos x = c has two solutions exactly when −1 < c < 0. Now cos² 2x = c² means cos 2x = |c| or cos 2x = −|c|, with 0 < |c| < 1. Putting t = 2x turns the range into 0 < t < 3π, which covers one full period plus a half. Each of the two values gives 2 solutions in (0, 2π) and 1 more in (2π,  3π), so 3 each, and the two sets are disjoint because |c| ≠ −|c|. That is 6 solutions.