O-Level A-Math 2020 Paper 1, Question 10: Solve cos 2x = k in range

(i) At the intersections \(\cos 2x = -\dfrac{\sqrt3}{2}\). With \(0 \leqslant x \leqslant \pi\) we have \(0 \leqslant 2x \leqslant 2\pi\), and the basic angle is \(\dfrac{\pi}{6}\). As the cosine is negative, \(2x\) is in the 2nd or 3rd quadrant: \[2x = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6} \quad\text{or}\quad 2x = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}, \quad\text{so}\quad x = \dfrac{5\pi}{12} \ (A) \quad\text{or}\quad x = \dfrac{7\pi}{12} \ (B).\]

(ii) Between \(A\) and \(B\) the curve lies below the line, so the shaded area is \[\text{Area} = \int_{5\pi/12}^{7\pi/12} \left[ -\dfrac{\sqrt3}{2} - \cos 2x \right] dx = \left[ -\dfrac{\sqrt3}{2}\,x - \dfrac{\sin 2x}{2} \right]_{5\pi/12}^{7\pi/12}.\] At \(x = \dfrac{7\pi}{12}\): \(-\dfrac{\sqrt3}{2}\cdot\dfrac{7\pi}{12} - \dfrac{\sin\frac{7\pi}{6}}{2} = -\dfrac{7\sqrt3\,\pi}{24} + \dfrac{1}{4}\) (since \(\sin\frac{7\pi}{6} = -\tfrac12\)). At \(x = \dfrac{5\pi}{12}\): \(-\dfrac{\sqrt3}{2}\cdot\dfrac{5\pi}{12} - \dfrac{\sin\frac{5\pi}{6}}{2} = -\dfrac{5\sqrt3\,\pi}{24} - \dfrac{1}{4}\) (since \(\sin\frac{5\pi}{6} = \tfrac12\)). Subtracting, \[\text{Area} = \left(-\dfrac{7\sqrt3\,\pi}{24} + \dfrac14\right) - \left(-\dfrac{5\sqrt3\,\pi}{24} - \dfrac14\right) = \dfrac{1}{2} - \dfrac{2\sqrt3\,\pi}{24} = \dfrac{1}{2} - \dfrac{\pi\sqrt{3}}{12} \ \text{units}^2.\] (Equivalently: rectangle from the \(x\)-axis down to the line over width \(\tfrac{\pi}{6}\) is \(\tfrac{\pi}{6}\cdot\tfrac{\sqrt3}{2} = \tfrac{\pi\sqrt3}{12}\); the area between the curve and the \(x\)-axis from \(A\) to \(B\) is \(\left|\int_{5\pi/12}^{7\pi/12}\cos 2x\,dx\right| = \tfrac12\); the shaded lens is the difference, \(\tfrac12 - \tfrac{\pi\sqrt3}{12}\).)